How to Combine Limiting-Reagent and Percentage-Yield Calculations
Learn how to identify the limiting reagent, calculate theoretical yield and leftover reactants, then work out percentage yield. Includes step-by-step HSC Chemistry examples, contrasting scenarios, and challenging practice questions with worked solutions.
A gas syringe collects 0.995 L of hydrogen when 3.00 g of magnesium reacts with 100.0 mL of 1.00 mol L⁻¹ hydrochloric acid. If you calculate the hydrogen produced from the magnesium alone, you predict about 3.06 L. So why is the measured volume much smaller?
The acid runs out first. Once it has been consumed, the remaining magnesium cannot produce any more hydrogen.
This is where two important calculations work together. **Limiting-reagent calculations determine the theoretical yield, while percentage yield compares that theoretical amount with the product actually collected.** You need to complete them in that order.
The combined method is:
1. Balance the chemical equation.
2. Convert every reactant into moles.
3. Divide each reactant's mole amount by its coefficient.
4. Identify the smallest result as the limiting reagent.
5. Use the limiting reagent to calculate the theoretical product amount.
6. Calculate the amount of excess reactant remaining.
7. Compare the actual and theoretical yields to calculate percentage yield.
Let's build the method carefully, including what happens when the experiment doesn't collect all the product.
## 1. Start with the balanced equation
Consider the reaction between magnesium and hydrochloric acid:
\[
\ce{Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)}
\]
The coefficients tell us that:
- 1 mol of magnesium reacts with 2 mol of hydrochloric acid.
- 1 mol of magnesium produces 1 mol of hydrogen gas.
- 2 mol of hydrochloric acid produce 1 mol of hydrogen gas.
These ratios determine how much product can form. The masses and volumes you start with tell you how many moles are available, but they don't tell you which reactant runs out first.
### Why divide by the coefficient?
Suppose a reaction follows:
\[
\ce{2A + 3B -> 2C}
\]
You have 0.80 mol of A and 0.90 mol of B.
A common mistake is to choose B because it has more moles, or A because it has fewer. Neither comparison is reliable.
Instead, divide each amount by its coefficient:
| Reactant | Initial amount | Coefficient | Amount ÷ coefficient |
|---|---:|---:|---:|
| A | 0.80 mol | 2 | 0.40 |
| B | 0.90 mol | 3 | 0.30 |
B is the limiting reagent because its result is smaller.
Think of each quotient as the number of reaction batches that a reactant can support. A can support 0.40 batches, but B can support only 0.30. Once B runs out, the reaction must stop, even though A remains.
The general rule is:
\[
\boxed{\text{Limiting reagent}
=\text{smallest}\left(\frac{n}{\text{coefficient}}\right)}
\]
Here, \(n\) is the initial amount of a reactant in moles.
This comparison only works with a balanced equation. If the coefficients are wrong, your limiting-reagent decision will be wrong too.
For a more detailed explanation of identifying limiting reagents, see [Limiting Reagent Reactions: HSC Chemistry Explained](https://tutorgum.com/learn/resources/hsc-chemistry-limiting-reagent-reactions/).
## 2. Calculate the theoretical yield
Once you know the limiting reagent, use its mole amount to calculate the maximum possible amount of product.
This maximum is called the **theoretical yield**.
For the general reaction:
\[
\ce{2A + 3B -> 2C}
\]
if B is limiting, the product amount is:
\[
n(C)=n(B)\times\frac{2}{3}
\]
The numerator comes from the product coefficient, while the denominator comes from the limiting-reagent coefficient.
You can use the same approach to calculate a theoretical mass:
\[
m=nM
\]
where:
- \(m\) is mass in grams (g);
- \(n\) is amount in moles (mol);
- \(M\) is molar mass in grams per mole (g mol⁻¹).
If the product is a gas, you may instead calculate its volume using the appropriate molar volume or gas equation.
The important point is that you calculate theoretical product from the limiting reagent, not whichever reactant happens to have the larger mass or mole amount.
## 3. Calculate how much excess reactant remains
Finding the limiting reagent doesn't finish the calculation. You may also need to determine how much of the excess reagent is left after the reaction.
The method is:
\[
\boxed{
n_{\text{remaining}}
=
n_{\text{initial}}-n_{\text{reacted}}
}
\]
The balanced equation tells you how much of the excess reactant must react with the limiting reagent.
For the general reaction:
\[
\ce{2A + 3B -> 2C}
\]
if B is limiting, the amount of A consumed is:
\[
n(A)_{\text{reacted}}
=
n(B)_{\text{initial}}\times\frac{2}{3}
\]
Then subtract this amount from the initial amount of A.
There is an important assumption here: this calculation assumes the limiting reagent reacts completely. If a reaction stops before that happens, some limiting reagent may remain as well. A percentage yield by itself cannot tell you exactly how much reactant remains.
## 4. Calculate percentage yield
Theoretical yield tells you what should be produced under ideal conditions. Actual yield is the amount of product obtained experimentally.
In a real experiment, the actual yield may be lower because:
- some reactants do not react completely;
- competing reactions consume reactants;
- product is lost during transfer or filtration;
- some product remains dissolved in the solution;
- the collection or measurement process is imperfect.
Percentage yield measures the relationship between these two amounts:
\[
\boxed{
\%\text{ yield}
=
\frac{\text{Actual yield}}
{\text{Theoretical yield}}
\times100
}
\]
You must compare equivalent quantities. For example, compare actual mass with theoretical mass, or actual moles with theoretical moles.
If both gas volumes are measured at the same temperature and pressure, you can also compare their volumes directly.
<text></text>
A percentage yield of 80% means that the amount of product collected was 80% of the theoretical maximum. It does not automatically mean that 20% of each reactant remains.
That distinction becomes particularly important when calculating excess reactant.
## 5. Understand what the experiment actually measures
Consider how the magnesium and hydrochloric acid reaction might be investigated.
A conical flask contains magnesium and hydrochloric acid. A bung seals the flask, and a delivery tube connects it to a gas syringe. Hydrogen travels through the tube and pushes the syringe plunger outwards.
For an accurate measurement, the reactants can be kept separate inside the flask until the bung and gas syringe are connected. The flask is then tilted to mix them.
[IMAGE: Annotated experimental apparatus showing a sealed conical flask containing a magnesium strip and hydrochloric acid initially held separately in a small inner vial. A gas-tight bung connects the flask to a horizontal gas syringe through a delivery tube. Show hydrogen moving from the flask into the syringe and label magnesium, hydrochloric acid, hydrogen, delivery tube, and gas syringe.]
What should this apparatus measure?
If the limiting reagent reacts completely and all hydrogen is collected, the measured gas amount should approach the theoretical yield.
An initially tempting design would be to leave the flask open on a balance and calculate hydrogen production from the decrease in mass. However, this is not a selective measurement of hydrogen. Water evaporation, splashing, and liquid droplets can also change the balance reading.
The gas syringe is a better measurement approach because it collects and measures the gas directly. However, it still has limitations. Leaks, incomplete collection, and temperature changes can affect the reading. Gas collected over water also requires a correction for water vapour if calculating the amount of dry hydrogen.
This is why an experimental percentage yield is a measured result, not simply a second theoretical calculation.
All numerical datasets in the following worked examples are illustrative.
### Worked example: Calculating copper yield and excess aluminium
**Question**
2.70 g of aluminium reacts with 13.5 g of anhydrous copper(II) chloride according to:
\[
\ce{2Al + 3CuCl2 -> 2AlCl3 + 3Cu}
\]
After the reaction, 5.10 g of dry copper is collected.
Calculate:
- the limiting reagent;
- the theoretical mass of copper;
- the mass of aluminium remaining;
- the percentage yield of copper.
Use the following molar masses:
- \(M(\ce{Al})=26.98\text{ g mol}^{-1}\)
- \(M(\ce{CuCl2})=134.45\text{ g mol}^{-1}\)
- \(M(\ce{Cu})=63.55\text{ g mol}^{-1}\)
Assume the limiting reagent reacts completely, there are no competing reactions, and the collected copper is pure and dry.
[IMAGE: Experimental arrangement showing a beaker containing copper(II) chloride solution and an aluminium strip, with reddish-brown copper forming. Beside it, show the filtration funnel separating the copper solid from the solution, followed by a drying dish containing the isolated copper on a balance. Label aluminium, copper(II) chloride solution, copper residue, filtrate, and balance.]
**Step 1**: Convert both reactants into moles.
Use:
\[
n=\frac{m}{M}
\]
For aluminium:
\[
n(\ce{Al})=\frac{2.70}{26.98}
=0.1001\text{ mol}
\]
For copper(II) chloride:
\[
n(\ce{CuCl2})=\frac{13.5}{134.45}
=0.1004\text{ mol}
\]
**Step 2**: Divide each mole amount by its coefficient.
The balanced equation is:
\[
\ce{2Al + 3CuCl2 -> 2AlCl3 + 3Cu}
\]
Therefore:
\[
\frac{n(\ce{Al})}{2}
=\frac{0.1001}{2}
=0.05005
\]
\[
\frac{n(\ce{CuCl2})}{3}
=\frac{0.1004}{3}
=0.03347
\]
The smaller result belongs to \(\ce{CuCl2}\).
Therefore, copper(II) chloride is the limiting reagent, and aluminium is in excess.
Notice the slightly surprising result: copper(II) chloride actually has more moles than aluminium. Comparing raw mole amounts would give you the wrong answer.
**Step 3**: Calculate the theoretical amount of copper.
From the balanced equation:
\[
3\text{ mol }\ce{CuCl2}
:
3\text{ mol }\ce{Cu}
\]
The mole ratio is 1:1.
Therefore:
\[
n(\ce{Cu})=0.1004\times\frac{3}{3}
=0.1004\text{ mol}
\]
Convert this into mass:
\[
\begin{aligned}
m(\ce{Cu})&=nM\\
&=0.1004\times63.55\\
&=6.38\text{ g}
\end{aligned}
\]
The theoretical yield is **6.38 g of copper**.
**Step 4**: Calculate the aluminium remaining.
The equation requires 2 mol of aluminium for every 3 mol of copper(II) chloride.
Therefore:
\[
n(\ce{Al})_{\text{reacted}}
=0.1004\times\frac{2}{3}
\]
\[
=0.06694\text{ mol}
\]
Subtract the amount reacted from the initial amount:
\[
\begin{aligned}
n(\ce{Al})_{\text{remaining}}
&=0.1001-0.06694\\
&=0.0331\text{ mol}
\end{aligned}
\]
Convert back into mass:
\[
\begin{aligned}
m(\ce{Al})_{\text{remaining}}
&=0.0331\times26.98\\
&=0.894\text{ g}
\end{aligned}
\]
Therefore, approximately 0.894 g of aluminium remains.
**Step 5**: Calculate percentage yield.
We know:
- Actual yield = 5.10 g
- Theoretical yield = 6.38 g
Substitute:
\[
\begin{aligned}
\%\text{ yield}
&=\frac{5.10}{6.38}\times100\\
&=79.9\%
\end{aligned}
\]
The final answers are:
| Quantity | Result |
|---|---:|
| Limiting reagent | \(\ce{CuCl2}\) |
| Excess reagent | \(\ce{Al}\) |
| Theoretical copper yield | 6.38 g |
| Aluminium remaining | 0.894 g |
| Percentage yield | 79.9% |
**Why this calculation matters experimentally**
The copper must be separated from the remaining reaction mixture, washed, and dried before its mass is used as the actual yield.
If you weigh it while wet, the retained water increases the measured mass. If copper is lost during filtration or transfer, the measured mass decreases.
The theoretical yield does not change just because some product is lost during collection.
There is one qualification: our calculation of 0.894 g of remaining aluminium assumes all the limiting reagent has reacted. If the reaction stopped prematurely, some copper(II) chloride could also remain. You would need additional experimental information to calculate the actual leftover amounts.
### Worked example: Combining concentration, gas volume, leftovers, and percentage yield
**Question**
3.00 g of magnesium reacts with 100.0 mL of 1.00 mol L⁻¹ hydrochloric acid.
The balanced equation is:
\[
\ce{Mg + 2HCl -> MgCl2 + H2}
\]
A gas syringe collects 0.995 L of dry hydrogen at 25.0°C and 100 kPa.
Calculate:
- the limiting reagent;
- the theoretical volume of hydrogen;
- the mass of magnesium remaining;
- the percentage yield of hydrogen.
Use:
- \(M(\ce{Mg})=24.31\text{ g mol}^{-1}\)
- Molar volume at the specified conditions = \(24.8\text{ L mol}^{-1}\).
Assume no competing reactions and complete consumption of the limiting reagent. The measured gas volume has already been corrected to dry hydrogen at the stated temperature and pressure.
**Step 1**: Convert both reactants into moles.
For magnesium:
\[
\begin{aligned}
n(\ce{Mg})&=\frac{m}{M}\\
&=\frac{3.00}{24.31}\\
&=0.1234\text{ mol}
\end{aligned}
\]
For hydrochloric acid, use:
\[
n=cV
\]
where \(c\) is concentration in mol L⁻¹ and \(V\) is volume in litres.
First convert the volume:
\[
100.0\text{ mL}=0.1000\text{ L}
\]
Then:
\[
\begin{aligned}
n(\ce{HCl})&=1.00\times0.1000\\
&=0.1000\text{ mol}
\end{aligned}
\]
**Step 2**: Identify the limiting reagent.
The equation is:
\[
\ce{Mg + 2HCl -> MgCl2 + H2}
\]
Divide each mole amount by its coefficient:
\[
\frac{n(\ce{Mg})}{1}
=\frac{0.1234}{1}
=0.1234
\]
\[
\frac{n(\ce{HCl})}{2}
=\frac{0.1000}{2}
=0.0500
\]
The smaller result belongs to hydrochloric acid.
Therefore, \(\ce{HCl}\) is the limiting reagent.
**Step 3**: Calculate the theoretical hydrogen volume.
The equation shows:
\[
2\text{ mol }\ce{HCl}
:
1\text{ mol }\ce{H2}
\]
Therefore:
\[
n(\ce{H2})=0.1000\times\frac{1}{2}
\]
\[
=0.0500\text{ mol}
\]
Use the molar volume:
\[
V=nV_m
\]
\[
\begin{aligned}
V(\ce{H2})&=0.0500\times24.8\\
&=1.24\text{ L}
\end{aligned}
\]
The theoretical hydrogen yield is 1.24 L.
**Step 4**: Calculate the magnesium remaining.
The equation shows that 1 mol of magnesium reacts with 2 mol of hydrochloric acid.
Therefore:
\[
n(\ce{Mg})_{\text{reacted}}
=0.1000\times\frac{1}{2}
=0.0500\text{ mol}
\]
The amount of magnesium remaining is:
\[
\begin{aligned}
n(\ce{Mg})_{\text{remaining}}
&=0.1234-0.0500\\
&=0.0734\text{ mol}
\end{aligned}
\]
Convert this into mass:
\[
\begin{aligned}
m(\ce{Mg})_{\text{remaining}}
&=0.0734\times24.31\\
&=1.78\text{ g}
\end{aligned}
\]
Therefore, 1.78 g of magnesium remains.
**Step 5**: Calculate percentage yield.
The actual hydrogen volume is 0.995 L.
The theoretical hydrogen volume is 1.24 L.
Since both volumes are measured at the same temperature and pressure, we can compare them directly:
\[
\begin{aligned}
\%\text{ yield}
&=\frac{0.995}{1.24}\times100\\
&=80.2\%
\end{aligned}
\]
The final answers are:
| Quantity | Result |
|---|---:|
| Limiting reagent | \(\ce{HCl}\) |
| Excess reagent | \(\ce{Mg}\) |
| Theoretical hydrogen volume | 1.24 L |
| Magnesium remaining | 1.78 g |
| Percentage yield | 80.2% |
Notice that the theoretical yield is determined by the amount of hydrochloric acid, while the actual yield comes from the gas syringe measurement.
If the two gas volumes were measured at different temperatures or pressures, you could not simply divide one by the other. You would first need to convert them to equivalent conditions or calculate the corresponding mole amounts.
## 7. The difference between theoretical leftovers and actual leftovers
This is one of the most important distinctions when combining these calculations.
The usual limiting-reagent method assumes the limiting reagent reacts completely. Under that assumption, you can calculate the amount of excess reagent remaining.
However, a low percentage yield does not necessarily mean the reaction has stopped with excess reagent still present.
Consider two possibilities:
| Situation | What happens? | What can you conclude? |
|---|---|---|
| Product is lost during filtration | Reactants may have reacted completely, but some product is not collected. | The theoretical leftover calculation remains valid. |
| Reaction is incomplete | Some of the limiting reagent may remain unreacted. | You cannot assume the limiting reagent has been completely consumed. |
For example, suppose a reaction has a theoretical yield of 10.0 g, but only 8.00 g is collected.
The percentage yield is 80.0%.
That result alone cannot tell you whether:
- 2.00 g of product was lost during collection;
- the reaction stopped before all the limiting reagent reacted;
- a competing reaction consumed some reactant;
- or a combination of these occurred.
You need experimental evidence to distinguish these possibilities.
### Can percentage yield be greater than 100%?
A calculated yield above 100% is possible, but it should make you question the measurements or assumptions.
Common explanations include:
- the collected product contains water;
- the product is contaminated with unreacted material;
- the theoretical yield was underestimated;
- the starting reactant amounts or chemical equation were incorrect.
Do not automatically change the answer to 100%.
Instead, check whether the actual product mass represents the mass of pure, dry product.
There is another possibility that is particularly important in multi-stage reactions: the reactants supplied to the system may have changed. If additional reactant is added, the original theoretical yield is no longer the correct maximum for the entire process.
The final practice question explores this idea.
## Questions and solutions
The following questions progress from a standard limiting-reagent calculation to more demanding problems involving exact stoichiometric ratios, reactant purity, and changes to a reaction partway through.
### Question 1
Nitrogen reacts with hydrogen to form ammonia:
\[
\ce{N2 + 3H2 -> 2NH3}
\]
A reaction begins with 0.500 mol of nitrogen and 1.20 mol of hydrogen. At the end, 0.600 mol of pure ammonia is collected.
Calculate:
a) The limiting reagent.
b) The theoretical amount of ammonia.
c) The amount of nitrogen remaining.
d) The percentage yield of ammonia.
Assume the limiting reagent reacts completely and there are no competing reactions.
### Solution 1
**Hydrogen is the limiting reagent, the theoretical ammonia yield is 0.800 mol, 0.100 mol of nitrogen remains, and the percentage yield is 75.0%.**
**Step 1**: Divide each reactant amount by its coefficient.
For nitrogen:
\[
\frac{0.500}{1}=0.500
\]
For hydrogen:
\[
\frac{1.20}{3}=0.400
\]
Hydrogen has the smaller result, so it is limiting.
**Step 2**: Calculate the theoretical amount of ammonia.
From the equation:
\[
3\text{ mol }\ce{H2}
:
2\text{ mol }\ce{NH3}
\]
Therefore:
\[
n(\ce{NH3})=1.20\times\frac{2}{3}
\]
\[
=0.800\text{ mol}
\]
**Step 3**: Calculate the nitrogen remaining.
The equation requires 1 mol of nitrogen for every 3 mol of hydrogen.
\[
n(\ce{N2})_{\text{reacted}}
=1.20\times\frac{1}{3}
=0.400\text{ mol}
\]
Subtract from the initial amount:
\[
n(\ce{N2})_{\text{remaining}}
=0.500-0.400
\]
\[
=0.100\text{ mol}
\]
**Step 4**: Calculate percentage yield.
\[
\begin{aligned}
\%\text{ yield}
&=\frac{0.600}{0.800}\times100\\
&=75.0\%
\end{aligned}
\]
The reaction produces 75.0% of its theoretical ammonia yield.
### Question 2
Ammonia reacts with sulfuric acid to produce ammonium sulfate:
\[
\ce{2NH3 + H2SO4 -> (NH4)2SO4}
\]
A reaction uses:
- 3.40 g of ammonia;
- 14.7 g of sulfuric acid.
After the reaction, 10.6 g of dry ammonium sulfate is collected.
Calculate:
a) The limiting reagent.
b) The theoretical mass of ammonium sulfate.
c) The mass of excess reactant remaining.
d) The percentage yield.
Use:
- \(M(\ce{NH3})=17.03\text{ g mol}^{-1}\)
- \(M(\ce{H2SO4})=98.08\text{ g mol}^{-1}\)
- \(M(\ce{(NH4)2SO4})=132.14\text{ g mol}^{-1}\)
Assume the limiting reagent reacts completely and the product is pure and dry.
### Solution 2
**Ammonia is the limiting reagent, the theoretical ammonium sulfate yield is 13.2 g, approximately 4.91 g of sulfuric acid remains, and the percentage yield is 80.4%.**
**Step 1**: Calculate the initial moles of each reactant.
For ammonia:
\[
n(\ce{NH3})=\frac{3.40}{17.03}
=0.1996\text{ mol}
\]
For sulfuric acid:
\[
n(\ce{H2SO4})=\frac{14.7}{98.08}
=0.1499\text{ mol}
\]
**Step 2**: Divide by the coefficients.
\[
\frac{n(\ce{NH3})}{2}
=\frac{0.1996}{2}
=0.0998
\]
\[
\frac{n(\ce{H2SO4})}{1}
=0.1499
\]
Ammonia has the smaller result, so it is limiting.
Notice that ammonia actually has more moles than sulfuric acid. The coefficients are what determine which reactant runs out first.
**Step 3**: Calculate the theoretical ammonium sulfate yield.
The equation gives:
\[
2\text{ mol }\ce{NH3}
:
1\text{ mol }\ce{(NH4)2SO4}
\]
Therefore:
\[
n(\ce{(NH4)2SO4})
=0.1996\times\frac{1}{2}
\]
\[
=0.0998\text{ mol}
\]
Convert to mass:
\[
\begin{aligned}
m&=nM\\
&=0.0998\times132.14\\
&=13.19\text{ g}
\end{aligned}
\]
The theoretical yield is 13.2 g.
**Step 4**: Calculate the sulfuric acid remaining.
The equation requires 1 mol of sulfuric acid for every 2 mol of ammonia.
Therefore:
\[
n(\ce{H2SO4})_{\text{reacted}}
=0.1996\times\frac{1}{2}
\]
\[
=0.0998\text{ mol}
\]
Subtract:
\[
\begin{aligned}
n(\ce{H2SO4})_{\text{remaining}}
&=0.1499-0.0998\\
&=0.0501\text{ mol}
\end{aligned}
\]
Convert to mass:
\[
\begin{aligned}
m&=0.0501\times98.08\\
&=4.91\text{ g}
\end{aligned}
\]
**Step 5**: Calculate percentage yield.
\[
\begin{aligned}
\%\text{ yield}
&=\frac{10.6}{13.19}\times100\\
&=80.4\%
\end{aligned}
\]
The experiment produces 80.4% of the theoretical ammonium sulfate yield.
### Question 3
Sodium hydroxide reacts with sulfuric acid:
\[
\ce{2NaOH(aq) + H2SO4(aq) -> Na2SO4(aq) + 2H2O(l)}
\]
A student mixes:
- 0.600 mol of sodium hydroxide;
- 0.300 mol of sulfuric acid.
The reaction is confirmed to have gone to completion. After isolating and drying the sodium sulfate, the student collects 0.240 mol of product.
The student claims:
"Since only 80% of the theoretical product was collected, 20% of both reactants must still be present."
Is this conclusion correct?
Calculate the theoretical amount of sodium sulfate and its percentage yield. Explain what happened to the reactants and why the student's interpretation is incorrect.
### Solution 3
**The student's conclusion is incorrect. Both reactants are present in exactly the required stoichiometric ratio, so neither is in excess. The theoretical sodium sulfate yield is 0.300 mol, and the percentage yield is 80.0%.**
**Step 1**: Compare the reactants using their coefficients.
For sodium hydroxide:
\[
\frac{0.600}{2}=0.300
\]
For sulfuric acid:
\[
\frac{0.300}{1}=0.300
\]
The results are equal.
This means both reactants are present in the exact ratio required by the balanced equation.
Neither reactant is in excess. If the reaction goes to completion, both are completely consumed.
**Step 2**: Calculate the theoretical amount of sodium sulfate.
The equation shows:
\[
1\text{ mol }\ce{H2SO4}
:
1\text{ mol }\ce{Na2SO4}
\]
Therefore:
\[
n(\ce{Na2SO4})=0.300\text{ mol}
\]
This is the theoretical yield.
**Step 3**: Calculate percentage yield.
\[
\begin{aligned}
\%\text{ yield}
&=\frac{0.240}{0.300}\times100\\
&=80.0\%
\end{aligned}
\]
The isolated product represents 80.0% of the theoretical yield.
**Step 4**: Explain the missing product.
The reaction was confirmed to have gone to completion, so the reactants were consumed.
The difference between 0.300 mol theoretically produced and 0.240 mol recovered is:
\[
0.300-0.240=0.060\text{ mol}
\]
This 0.060 mol represents product that was not recovered during isolation.
It does not represent unreacted sodium hydroxide or sulfuric acid.
This question exposes an important misconception: percentage yield measures the amount of product recovered relative to the theoretical maximum. It does not directly measure the amount of reactant remaining.
### Question 4
Iron can be produced by reducing iron(III) oxide with carbon monoxide:
\[
\ce{Fe2O3 + 3CO -> 2Fe + 3CO2}
\]
An illustrative sample of iron ore has a mass of 20.0 g and contains 80.0% iron(III) oxide by mass. The remaining 20.0% consists of inert impurities that do not react.
The sample reacts with 10.0 g of carbon monoxide. After the reaction, 9.20 g of pure, dry iron is collected.
Calculate:
a) The initial amount of iron(III) oxide.
b) The limiting reagent.
c) The theoretical mass of iron.
d) The mass of carbon monoxide remaining.
e) The percentage yield of iron.
Use:
- \(M(\ce{Fe2O3})=159.69\text{ g mol}^{-1}\)
- \(M(\ce{CO})=28.01\text{ g mol}^{-1}\)
- \(M(\ce{Fe})=55.85\text{ g mol}^{-1}\)
[IMAGE: Simplified cross-sectional illustration of a reaction chamber containing a central bed of iron(III) oxide ore. Carbon monoxide enters through one side and flows through the solid bed, while the iron product forms within the material and carbon dioxide leaves through the outlet. Show the inert gangue mixed with the ore. Label Fe2O3, CO entering, Fe product, inert impurities, and CO2 leaving.]
### Solution 4
**Iron(III) oxide is the limiting reagent. The theoretical iron yield is 11.2 g, approximately 1.58 g of carbon monoxide remains, and the percentage yield is 82.2%.**
**Step 1**: Calculate the mass of pure iron(III) oxide.
Only 80.0% of the ore is reactive iron(III) oxide.
\[
\begin{aligned}
m(\ce{Fe2O3})
&=20.0\times\frac{80.0}{100}\\
&=16.0\text{ g}
\end{aligned}
\]
The remaining 4.00 g is inert material and must not be included in the reactant mole calculation.
Convert the pure iron(III) oxide into moles:
\[
n(\ce{Fe2O3})=\frac{16.0}{159.69}
\]
\[
=0.1002\text{ mol}
\]
**Step 2**: Calculate the initial moles of carbon monoxide.
\[
n(\ce{CO})=\frac{10.0}{28.01}
\]
\[
=0.3570\text{ mol}
\]
**Step 3**: Identify the limiting reagent.
Divide each amount by its coefficient:
\[
\frac{n(\ce{Fe2O3})}{1}
=0.1002
\]
\[
\frac{n(\ce{CO})}{3}
=\frac{0.3570}{3}
=0.1190
\]
The smaller result belongs to iron(III) oxide.
Therefore, \(\ce{Fe2O3}\) is limiting and carbon monoxide is in excess.
**Step 4**: Calculate the theoretical iron yield.
The equation shows:
\[
1\text{ mol }\ce{Fe2O3}
:
2\text{ mol }\ce{Fe}
\]
Therefore:
\[
n(\ce{Fe})=0.1002\times2
\]
\[
=0.2004\text{ mol}
\]
Convert to mass:
\[
\begin{aligned}
m(\ce{Fe})&=0.2004\times55.85\\
&=11.19\text{ g}
\end{aligned}
\]
The theoretical yield is approximately 11.2 g of iron.
**Step 5**: Calculate the carbon monoxide remaining.
The equation requires 3 mol of carbon monoxide for every 1 mol of iron(III) oxide.
\[
n(\ce{CO})_{\text{reacted}}
=0.1002\times3
\]
\[
=0.3006\text{ mol}
\]
Subtract from the initial amount:
\[
\begin{aligned}
n(\ce{CO})_{\text{remaining}}
&=0.3570-0.3006\\
&=0.0564\text{ mol}
\end{aligned}
\]
Convert to mass:
\[
\begin{aligned}
m(\ce{CO})_{\text{remaining}}
&=0.0564\times28.01\\
&=1.58\text{ g}
\end{aligned}
\]
**Step 6**: Calculate percentage yield.
The actual iron mass is 9.20 g.
The theoretical iron mass is 11.19 g.
\[
\begin{aligned}
\%\text{ yield}
&=\frac{9.20}{11.19}\times100\\
&=82.2\%
\end{aligned}
\]
The experiment produces 82.2% of its theoretical iron yield.
The important extra step in this question was correcting for the purity of the ore before calculating moles. Using the entire 20.0 g as though it were pure iron(III) oxide would overestimate the theoretical yield and give incorrect results throughout the calculation.
### Question 5
Consider the reaction:
\[
\ce{2Mg + O2 -> 2MgO}
\]
A reaction vessel initially contains:
- 1.20 mol of magnesium;
- 0.500 mol of oxygen.
After the first reaction stage, 0.800 mol of pure, dry magnesium oxide is recovered. Analysis of the remaining mixture shows that it contains:
- 0.400 mol of unreacted magnesium;
- 0.100 mol of unreacted oxygen.
The magnesium oxide is separated from the remaining mixture. A further 0.600 mol of oxygen is then added to the vessel.
After the second stage, the total amount of magnesium oxide recovered across both stages is 1.14 mol.
Assume:
- no magnesium is added after the first stage;
- no side reactions occur;
- all measured magnesium oxide is pure;
- no product is lost during either collection;
- the measured remaining reactants and recovered product amounts are accurate.
Answer the following:
a) Identify the initial limiting reagent and calculate the theoretical magnesium oxide yield for the first stage.
b) Explain whether the first stage's 80% yield is caused by product loss or incomplete reaction. Verify your conclusion using the remaining reactant amounts.
c) After the additional oxygen is supplied, identify the limiting reagent and calculate the maximum additional amount of magnesium oxide that can form.
d) Calculate the percentage yield for the second stage and the overall percentage yield for the entire process.
e) Calculate the amounts of magnesium and oxygen remaining after the second stage.
A student calculates the overall percentage yield as 114% by dividing 1.14 mol by the original theoretical yield of 1.00 mol. Explain precisely why this calculation is invalid.
[IMAGE: Two-stage material-balance illustration for the reaction 2Mg + O2 -> 2MgO. Show the first vessel receiving 1.20 mol Mg and 0.500 mol O2, with 0.800 mol MgO separated and 0.400 mol Mg plus 0.100 mol O2 remaining. Show a second stage in which 0.600 mol additional O2 enters the remaining mixture, followed by collection of a total of 1.14 mol MgO across both stages. Preserve the supplied quantities, show the physical separation of product between stages, and do not label calculated limiting reagents or percentage yields.]
### Solution 5
**The initial limiting reagent is oxygen. After additional oxygen is supplied, magnesium becomes limiting. The correct overall theoretical yield is 1.20 mol of magnesium oxide, giving an overall percentage yield of 95.0%, not 114%.**
This question is more difficult because the available reactants change partway through the process. The theoretical yield must be recalculated when the starting conditions change.
**Step 1**: Identify the initial limiting reagent.
Initially:
\[
n(\ce{Mg})=1.20\text{ mol}
\]
\[
n(\ce{O2})=0.500\text{ mol}
\]
Divide by the coefficients:
\[
\frac{n(\ce{Mg})}{2}
=\frac{1.20}{2}
=0.600
\]
\[
\frac{n(\ce{O2})}{1}
=0.500
\]
Oxygen has the smaller result, so it is initially limiting.
**Step 2**: Calculate the first theoretical yield.
From the balanced equation:
\[
1\text{ mol }\ce{O2}
:
2\text{ mol }\ce{MgO}
\]
Therefore:
\[
n(\ce{MgO})_{\text{theoretical}}
=0.500\times2
\]
\[
=1.00\text{ mol}
\]
The first stage has a theoretical yield of 1.00 mol of magnesium oxide.
**Step 3**: Calculate the first-stage percentage yield and investigate the leftovers.
The actual amount recovered is 0.800 mol.
\[
\begin{aligned}
\%\text{ yield}
&=\frac{0.800}{1.00}\times100\\
&=80.0\%
\end{aligned}
\]
However, we also have information about the unreacted materials.
If 0.800 mol of \(\ce{MgO}\) forms, the equation tells us that:
\[
n(\ce{Mg})_{\text{reacted}}=0.800\text{ mol}
\]
\[
n(\ce{O2})_{\text{reacted}}=0.400\text{ mol}
\]
Check the remaining amounts.
For magnesium:
\[
1.20-0.800=0.400\text{ mol}
\]
For oxygen:
\[
0.500-0.400=0.100\text{ mol}
\]
These match the measured leftovers exactly.
This tells us that both reactants remain after the first stage.
The reaction stopped before the available limiting reagent was completely consumed. Therefore, the 80.0% yield in this example reflects incomplete reaction, not product lost during collection.
This is a crucial distinction: a percentage yield alone cannot establish why the yield is below 100%, but the measured leftover reactants provide additional evidence.
**Step 4**: Identify the limiting reagent after adding oxygen.
Now 0.600 mol of additional oxygen is added.
The remaining mixture contains:
- 0.400 mol of magnesium;
- 0.100 mol of oxygen.
After adding oxygen:
\[
n(\ce{O2})=0.100+0.600
\]
\[
=0.700\text{ mol}
\]
The new starting amounts for the second stage are therefore:
\[
n(\ce{Mg})=0.400\text{ mol}
\]
\[
n(\ce{O2})=0.700\text{ mol}
\]
Divide by the coefficients:
\[
\frac{n(\ce{Mg})}{2}
=\frac{0.400}{2}
=0.200
\]
\[
\frac{n(\ce{O2})}{1}
=0.700
\]
Magnesium now has the smaller result.
Therefore, magnesium becomes the limiting reagent in the second stage.
Notice that the limiting reagent has changed. Oxygen limited the first stage, but the additional oxygen makes magnesium the limiting reagent in the second stage.
**Step 5**: Calculate the maximum additional magnesium oxide.
The equation shows:
\[
2\text{ mol }\ce{Mg}
:
2\text{ mol }\ce{MgO}
\]
Therefore:
\[
n(\ce{MgO})_{\text{additional,theoretical}}
=0.400\times\frac{2}{2}
\]
\[
=0.400\text{ mol}
\]
The maximum additional product is 0.400 mol of magnesium oxide.
**Step 6**: Calculate the second-stage percentage yield.
The total product collected across both stages is 1.14 mol.
The first stage produced 0.800 mol, so the second stage produced:
\[
n(\ce{MgO})_{\text{additional,actual}}
=1.14-0.800
\]
\[
=0.340\text{ mol}
\]
Therefore:
\[
\begin{aligned}
\%\text{ yield}_{\text{stage 2}}
&=\frac{0.340}{0.400}\times100\\
&=85.0\%
\end{aligned}
\]
The second-stage percentage yield is 85.0%.
**Step 7**: Recalculate the overall theoretical yield.
This is the critical part of the question.
The original theoretical yield of 1.00 mol applied only to the original quantities:
- 1.20 mol Mg;
- 0.500 mol O2.
But an additional 0.600 mol of oxygen was subsequently supplied.
The total amounts of reactants supplied across the entire process are now:
\[
n(\ce{Mg})_{\text{total}}=1.20\text{ mol}
\]
\[
n(\ce{O2})_{\text{total}}=0.500+0.600
\]
\[
=1.10\text{ mol}
\]
Compare these new amounts:
\[
\frac{1.20}{2}=0.600
\]
\[
\frac{1.10}{1}=1.10
\]
Magnesium is the limiting reagent for the overall process.
The updated theoretical magnesium oxide yield is:
\[
n(\ce{MgO})_{\text{theoretical}}
=1.20\times\frac{2}{2}
\]
\[
=1.20\text{ mol}
\]
Now calculate the overall percentage yield:
\[
\begin{aligned}
\%\text{ yield}_{\text{overall}}
&=\frac{1.14}{1.20}\times100\\
&=95.0\%
\end{aligned}
\]
The correct overall percentage yield is 95.0%.
Why isn't it 114%?
Because the original theoretical yield of 1.00 mol no longer represents the maximum possible product after the additional oxygen has been supplied.
Dividing 1.14 by 1.00 compares the final product with an outdated theoretical maximum. It ignores the extra reactant added between stages.
The denominator must reflect all reactants supplied to the process being assessed.
**Step 8**: Calculate the final remaining amounts.
We can now check the final material balance.
The total amount of magnesium oxide formed is 1.14 mol.
From the equation:
\[
2\text{ mol }\ce{Mg}
:
2\text{ mol }\ce{MgO}
\]
Therefore, magnesium consumed is:
\[
n(\ce{Mg})_{\text{reacted}}=1.14\text{ mol}
\]
The amount remaining is:
\[
1.20-1.14=0.060\text{ mol}
\]
For oxygen:
\[
1\text{ mol }\ce{O2}
:
2\text{ mol }\ce{MgO}
\]
Therefore:
\[
n(\ce{O2})_{\text{reacted}}
=\frac{1.14}{2}
=0.570\text{ mol}
\]
The total oxygen supplied was 1.10 mol.
Therefore:
\[
n(\ce{O2})_{\text{remaining}}
=1.10-0.570
\]
\[
=0.530\text{ mol}
\]
The final results are:
| Quantity | Result |
|---|---:|
| Initial limiting reagent | Oxygen |
| First-stage theoretical yield | 1.00 mol MgO |
| First-stage percentage yield | 80.0% |
| Second-stage limiting reagent | Magnesium |
| Maximum additional MgO | 0.400 mol |
| Second-stage percentage yield | 85.0% |
| Overall theoretical yield | 1.20 mol MgO |
| Overall percentage yield | 95.0% |
| Final magnesium remaining | 0.060 mol |
| Final oxygen remaining | 0.530 mol |
The central lesson is that theoretical yield depends on the reactants available for the particular reaction or process you are calculating. If the starting quantities change, you must reconsider the limiting reagent and recalculate the theoretical yield.
## 8. A quick method to remember for HSC questions
When a question combines limiting reagents, excess reactants, and percentage yield, keep the calculations in this order.
| Stage | What you do | Calculation |
|---|---|---|
| 1. Initial amounts | Convert reactants into moles | \(n=m/M\) or \(n=cV\) |
| 2. Limiting reagent | Divide each mole amount by its coefficient | \(n/\text{coefficient}\) |
| 3. Theoretical yield | Use the limiting reagent's mole ratio | Limiting reagent to product |
| 4. Excess remaining | Calculate consumption, then subtract | Initial moles − reacted moles |
| 5. Actual yield | Use the experimentally collected product | Given or measured amount |
| 6. Percentage yield | Compare actual with theoretical | \(\frac{\text{actual}}{\text{theoretical}}\times100\) |
Three checks will catch most errors:
- **Check the limiting reagent:** Have you divided by the balanced equation's coefficients rather than comparing raw mole amounts?
- **Check the leftovers:** Have you subtracted the amount that reacts, rather than the amount of product collected?
- **Check the yield:** Does your actual yield use the same units as your theoretical yield, and does the theoretical calculation use the correct starting quantities?
If your reactants are supplied as solutions, start by calculating their mole amounts using \(n=cV\), with volume in litres. The rest of the method is unchanged. For a worked precipitation example involving two solutions, see [Mixing Two Solutions: A Worked Limiting-Reagent Problem](https://tutorgum.com/learn/resources/mixing-two-solutions-a-worked-limiting-reagent-precipitation-problem/).
The key decision is always the same: establish what can react first, calculate what could theoretically form, and only then compare that prediction with what the experiment actually produced.