How to Calculate the Mass of Excess Reactant Left Over

Five grams of marble chips are added to hydrochloric acid. The mixture fizzes for a while, then the bubbling stops. There is still white solid sitting at the bottom of the flask.

Has all the marble reacted, or did the hydrochloric acid run out first?

This is exactly the problem you solve when calculating the mass of excess reactant left over. You need to work out which reactant limits the reaction, calculate how much of the other reactant was consumed, and subtract that amount from what you started with.

The key method is:

1. Calculate the initial number of moles of each reactant.
2. Divide each mole amount by its coefficient in the balanced equation.
3. Identify the reactant with the smallest result. This is the limiting reactant.
4. Use the mole ratio to calculate how much of the excess reactant was consumed.
5. Subtract the amount consumed from the initial amount.
6. Convert the remaining moles into mass.

Let’s use the marble and hydrochloric acid reaction to see why this works.

01A practical example: Which reactant runs out first?

Calcium carbonate is the main component of marble. It reacts with hydrochloric acid according to the balanced equation:

\[ \ce{CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l)} \]

Suppose we start with:

• 5.00 g of pure calcium carbonate, \(\ce{CaCO3}\).
• 50.0 mL of hydrochloric acid at a concentration of 1.00 mol L\(^{-1}\).

These are illustrative quantities.

The equation tells us that one mole of calcium carbonate reacts with two moles of hydrochloric acid.

First, calculate the initial amounts.

For calcium carbonate, using a molar mass of 100.09 g mol\(^{-1}\):

\[ n(\ce{CaCO3})=\frac{5.00}{100.09} \]

\[ =0.04996\text{ mol} \]

For hydrochloric acid, use \(n=cV\), remembering to convert millilitres into litres:

\[ n(\ce{HCl})=1.00\times0.0500 \]

\[ =0.0500\text{ mol} \]

Now divide each amount by its coefficient.

Hydrochloric acid has the smaller result, so it is the limiting reactant.

This means all the hydrochloric acid is consumed, but some calcium carbonate remains.

The amount of calcium carbonate consumed is:

\[ n(\ce{CaCO3\ consumed}) =\frac{0.0500}{2} \]

\[ =0.0250\text{ mol} \]

Subtract this from the initial amount:

\[ n(\ce{CaCO3\ left}) =0.04996-0.0250 \]

\[ =0.02496\text{ mol} \]

Finally, convert the remaining moles into mass:

\[ m=nM \]

\[ m=0.02496\times100.09 \]

\[ \boxed{m=2.50\text{ g}} \]

Approximately 2.50 g of calcium carbonate remains unreacted.

Notice that the initial amounts of the two reactants were quite similar. However, their coefficients were different. That is why comparing their raw mole amounts would not have been enough.

What would the experiment actually show?

The experimental question is whether the reaction stops because the marble has been consumed or because the acid has been used up.

A simple first attempt would be to add acid to marble chips in an open beaker and watch the bubbles.

The prediction is that carbon dioxide will form while the reaction proceeds. Eventually, the bubbling will stop.

But there is a problem: bubbling stopping does not tell us which reactant has run out. Both possibilities produce the same visible result.

A better design uses known quantities of both reactants and collects the gas in a gas syringe. After the reaction, the mixture can be filtered, the remaining solid washed, dried to constant mass, and weighed.

The stoichiometric calculation predicts 0.0250 mol of carbon dioxide. At 25°C and 100 kPa, using a molar gas volume of approximately 24.8 L mol\(^{-1}\), this corresponds to:

\[ V=0.0250\times24.8 \]

\[ V=0.620\text{ L} \]

This is an ideal-gas estimate. The actual gas syringe reading may be lower because some carbon dioxide dissolves in the solution or is lost through leaks.

The filtered, dried solid provides a separate measurement. For pure marble chips, the predicted remaining mass is approximately 2.50 g.

This is an illustrative experimental design, not a report of recorded measurements. Its purpose is to show how a theoretical stoichiometric prediction can be tested against an experimental result.

02Why do we divide the moles by the coefficient?

The coefficients in a balanced chemical equation represent the mole ratios required for a reaction.

Consider:

\[ \ce{N2 + 3H2 -> 2NH3} \]

One mole of nitrogen requires three moles of hydrogen.

Imagine each reaction as a small recipe. One complete reaction portion requires one mole of nitrogen and three moles of hydrogen.

The reactant that can supply the fewest complete reaction portions determines how far the reaction can proceed.

This is what dividing by the coefficient measures.

For a general reaction:

\[ \ce{aA + bB -> cC + dD} \]

the number of reaction portions available from each reactant is:

\[ \frac{n(A)}{a} \qquad\text{and}\qquad \frac{n(B)}{b} \]

The smaller value identifies the limiting reactant.

The analogy is useful for counting how much reaction can occur, but the equation does not mean molecules literally react in pre-arranged groups. It describes the overall mole ratio, not the detailed reaction mechanism.

A common mistake: comparing the initial moles directly

Suppose you have:

• 0.100 mol of nitrogen.
• 0.250 mol of hydrogen.

For the reaction:

\[ \ce{N2 + 3H2 -> 2NH3} \]

A student might look at the amounts and decide that nitrogen is limiting because there are fewer moles of nitrogen than hydrogen.

That prediction is wrong.

Divide each amount by its coefficient:

\[ \frac{n(\ce{N2})}{1}=0.100 \]

\[ \frac{n(\ce{H2})}{3}=0.0833 \]

Hydrogen is actually the limiting reactant.

The nitrogen consumed is:

\[ n(\ce{N2\ consumed})=\frac{0.250}{3} \]

\[ =0.0833\text{ mol} \]

The nitrogen remaining is:

\[ n(\ce{N2\ left})=0.100-0.0833 \]

\[ =0.0167\text{ mol} \]

Using a nitrogen molar mass of 28.02 g mol\(^{-1}\):

\[ m=0.0167\times28.02 \]

\[ \boxed{m=0.467\text{ g}} \]

Even though hydrogen started with more moles, it was the limiting reactant because the equation required three moles of hydrogen for every mole of nitrogen.

Always compare moles divided by coefficients, not just the initial mole amounts.

03A reliable method for calculating excess reactant

You can use the same process for almost any HSC stoichiometry question.

Step 1: Write the balanced chemical equation

Never start calculating before checking the equation.

An incorrect coefficient changes every subsequent mole ratio.

Step 2: Calculate the initial moles

Choose the appropriate formula.

For a reactant given as a mass:

\[ n=\frac{m}{M} \]

For a reactant in solution:

\[ n=cV \]

where:

• \(n\) is the amount in moles (mol).
• \(m\) is the mass in grams (g).
• \(M\) is the molar mass in g mol\(^{-1}\).
• \(c\) is the concentration in mol L\(^{-1}\).
• \(V\) is the solution volume in litres (L).

If the question gives a volume in millilitres, convert it into litres before using \(n=cV\).

Step 3: Divide each amount by its coefficient

For every reactant, calculate:

\[ \frac{\text{initial moles}}{\text{coefficient}} \]

The smallest value identifies the limiting reactant.

The other reactants are present in excess.

Step 4: Calculate how much of the excess reactant is consumed

Use the coefficients to establish the mole ratio.

For example, consider:

\[ \ce{2A + 3B -> products} \]

If B is limiting, the amount of A consumed is:

\[ n(A\text{ consumed})=\frac{2}{3}n(B\text{ initial}) \]

The coefficient ratio is essential. Do not assume that equal numbers of moles of the two reactants are consumed.

Step 5: Calculate the remaining moles

Subtract the amount consumed from the initial amount:

\[ n(\text{excess left}) = n(\text{excess initial}) - n(\text{excess consumed}) \]

Step 6: Convert the remaining moles into mass

If the question asks for the mass left over, finish with:

\[ \boxed{m(\text{excess left})=n(\text{excess left})M} \]

Your answer should be in grams if you have used a molar mass in g mol\(^{-1}\).

Step 7: Calculate the theoretical product yield if required

The limiting reactant also determines the maximum theoretical amount of product.

Use its mole amount and the product coefficient to calculate the theoretical product moles.

Then convert into mass if necessary.

If the question supplies an actual product mass, calculate percentage yield using:

\[ \%\text{ yield} = \frac{\text{actual yield}} {\text{theoretical yield}}\times100 \]

Keep these calculations separate. The theoretical amount of excess reactant remaining depends on the reaction stoichiometry, not on how much product was successfully collected.

04Worked examples

05Important checks and assumptions

Before moving on to practice questions, there are a few details worth checking in an exam.

1. Never subtract masses directly

If you start with 10 g of one reactant and 5 g of another, you cannot simply subtract 5 g from 10 g to find the excess.

The balanced equation gives mole ratios, not mass ratios.

Convert to moles first, apply the coefficients, and only then convert the remaining amount back into grams.

2. Be careful when the excess reactant is dissolved

Suppose hydrochloric acid is left over in an aqueous solution.

Using \(m=nM\) calculates the mass of hydrochloric acid remaining, not the total mass of the solution.

The solvent water is not included. You would need additional information, such as the solution density or the mass of the solvent, to calculate the total solution mass.

3. An impure reactant needs an extra calculation

If a question gives you an impure sample, calculate the mass of the actual reactive substance first.

For example, if a 20.0 g sample is 75.0% calcium carbonate:

\[ m(\ce{CaCO3})=20.0\times0.750 \]

\[ =15.0\text{ g} \]

Use 15.0 g when calculating the initial moles of calcium carbonate, not the full 20.0 g.

4. Check whether the reaction is assumed to go to completion

For standard stoichiometric calculations, assume the limiting reactant is completely consumed unless the question states otherwise.

A reversible reaction or a reaction that has been stopped early can leave both reactants present.

In that situation, the usual limiting-reactant calculation gives the maximum theoretical reaction extent, not necessarily the actual final amounts.

5. Check that your leftover amount makes sense

Your calculated leftover amount should never be negative.

If it is, check whether you identified the limiting reactant correctly or used the correct coefficient ratio.

Also check that the mass remaining cannot exceed the initial mass of that reactant.

06Questions and solutions

The following questions increase in difficulty. The first tests the basic method, while the later questions combine limiting reactants, purity, changing reaction conditions, and percentage yield.

Question 1

Magnesium reacts with oxygen to form magnesium oxide:

\[ \ce{2Mg(s) + O2(g) -> 2MgO(s)} \]

A reaction begins with 4.86 g of magnesium and 6.40 g of oxygen.

Assuming complete reaction, calculate:

a) Which reactant is limiting.

b) The mass of excess reactant remaining.

c) The theoretical mass of magnesium oxide produced.

Solution 1

The limiting reactant is magnesium. The reaction leaves 3.20 g of oxygen and produces a theoretical 8.06 g of magnesium oxide.

Step 1: Calculate the initial moles.

For magnesium, using \(M=24.3\text{ g mol}^{-1}\):

\[ n(\ce{Mg})=\frac{4.86}{24.3} =0.200\text{ mol} \]

For oxygen:

\[ n(\ce{O2})=\frac{6.40}{32.00} =0.200\text{ mol} \]

Step 2: Divide by the coefficients.

For magnesium:

\[ \frac{0.200}{2}=0.100 \]

For oxygen:

\[ \frac{0.200}{1}=0.200 \]

Magnesium has the smaller result, so it is limiting.

Step 3: Calculate the oxygen consumed.

The equation requires one mole of oxygen for every two moles of magnesium.

\[ n(\ce{O2\ consumed}) =0.200\times\frac{1}{2} \]

\[ =0.100\text{ mol} \]

Step 4: Calculate the oxygen remaining.

\[ n(\ce{O2\ left})=0.200-0.100 \]

\[ =0.100\text{ mol} \]

Convert into mass:

\[ m=0.100\times32.00 \]

\[ \boxed{m=3.20\text{ g}} \]

Step 5: Calculate the theoretical magnesium oxide yield.

The equation has a 2:2 mole ratio between magnesium and magnesium oxide.

Therefore:

\[ n(\ce{MgO})=0.200\text{ mol} \]

Using \(M(\ce{MgO})=40.3\text{ g mol}^{-1}\):

\[ m(\ce{MgO})=0.200\times40.3 \]

\[ \boxed{m(\ce{MgO})=8.06\text{ g}} \]

Magnesium is completely consumed, while oxygen remains in excess.

Question 2

Sodium hydrogen carbonate reacts with sulfuric acid according to:

\[ \ce{2NaHCO3(s) + H2SO4(aq) -> Na2SO4(aq) + 2CO2(g) + 2H2O(l)} \]

A reaction contains 16.8 g of sodium hydrogen carbonate and 11.8 g of pure sulfuric acid.

After the reaction, 12.0 g of sodium sulfate is recovered.

Calculate:

a) The limiting reactant.

b) The mass of excess reactant remaining.

c) The theoretical yield of sodium sulfate.

d) The percentage yield.

Explain why the actual product mass cannot be used directly to calculate the amount of excess acid remaining.

Solution 2

Sodium hydrogen carbonate is limiting. Approximately 1.99 g of sulfuric acid remains, the theoretical sodium sulfate yield is 14.2 g, and the percentage yield is 84.5%.

Step 1: Calculate the initial moles.

Using \(M(\ce{NaHCO3})=84.01\text{ g mol}^{-1}\):

\[ n(\ce{NaHCO3})=\frac{16.8}{84.01} \]

\[ =0.200\text{ mol} \]

For sulfuric acid, using \(M(\ce{H2SO4})=98.08\text{ g mol}^{-1}\):

\[ n(\ce{H2SO4})=\frac{11.8}{98.08} \]

\[ =0.1203\text{ mol} \]

Step 2: Divide each amount by its coefficient.

For sodium hydrogen carbonate:

\[ \frac{0.200}{2}=0.100 \]

For sulfuric acid:

\[ \frac{0.1203}{1}=0.1203 \]

Sodium hydrogen carbonate is limiting because it has the smaller result.

Sulfuric acid is in excess.

Step 3: Calculate the sulfuric acid consumed.

The equation requires two moles of sodium hydrogen carbonate for every mole of sulfuric acid.

\[ n(\ce{H2SO4\ consumed}) =\frac{0.200}{2} \]

\[ =0.100\text{ mol} \]

Step 4: Calculate the remaining sulfuric acid.

\[ n(\ce{H2SO4\ left}) =0.1203-0.100 \]

\[ =0.0203\text{ mol} \]

Convert to mass:

\[ m=0.0203\times98.08 \]

\[ \boxed{m=1.99\text{ g}} \]

Step 5: Calculate the theoretical sodium sulfate yield.

The mole ratio between sodium hydrogen carbonate and sodium sulfate is 2:1.

\[ n(\ce{Na2SO4})=\frac{0.200}{2} \]

\[ =0.100\text{ mol} \]

Using \(M(\ce{Na2SO4})=142.04\text{ g mol}^{-1}\):

\[ m=0.100\times142.04 \]

\[ \boxed{m=14.2\text{ g}} \]

Step 6: Calculate the percentage yield.

\[ \%\text{ yield}=\frac{12.0}{14.2}\times100 \]

\[ \boxed{\%\text{ yield}=84.5\%} \]

The actual yield is lower than the theoretical yield because not all the theoretically possible product was recovered.

This does not mean more sulfuric acid remained. Under the assumption of complete reaction, all the limiting sodium hydrogen carbonate was consumed, and the remaining acid is determined by the initial amounts and mole ratio.

Question 3

A limestone sample has a total mass of 25.0 g and contains 80.0% calcium carbonate by mass. The remaining material is an inert, insoluble impurity.

The sample reacts with 250.0 mL of hydrochloric acid at a concentration of 1.50 mol L\(^{-1}\).

The reaction is:

\[ \ce{CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l)} \]

Assume all the calcium carbonate reacts as far as the available acid allows.

Calculate:

a) The mass of excess calcium carbonate remaining.

b) The total mass of dry solid residue after filtration, washing, and drying.

c) The theoretical mass of carbon dioxide produced.

Explain why the total residue mass is not equal to the mass of excess reactant.

Solution 3

The excess calcium carbonate remaining is approximately 1.23 g. The total dry residue is 6.23 g, and the theoretical carbon dioxide production is 8.25 g.

Step 1: Calculate the actual mass of calcium carbonate.

Only 80.0% of the original sample is calcium carbonate.

\[ m(\ce{CaCO3})=25.0\times0.800 \]

\[ =20.0\text{ g} \]

The remaining 5.00 g is inert impurity.

Step 2: Calculate the initial moles of calcium carbonate.

\[ n(\ce{CaCO3})=\frac{20.0}{100.09} \]

\[ =0.1998\text{ mol} \]

Step 3: Calculate the initial moles of hydrochloric acid.

Convert the volume into litres:

\[ 250.0\text{ mL}=0.2500\text{ L} \]

Then:

\[ n(\ce{HCl})=1.50\times0.2500 \]

\[ =0.375\text{ mol} \]

Step 4: Identify the limiting reactant.

Divide each amount by its coefficient.

For calcium carbonate:

\[ \frac{0.1998}{1}=0.1998 \]

For hydrochloric acid:

\[ \frac{0.375}{2}=0.1875 \]

Hydrochloric acid is limiting.

Calcium carbonate is in excess.

Step 5: Calculate the calcium carbonate consumed.

The equation requires one mole of calcium carbonate for every two moles of hydrochloric acid.

\[ n(\ce{CaCO3\ consumed}) =\frac{0.375}{2} \]

\[ =0.1875\text{ mol} \]

Step 6: Calculate the calcium carbonate remaining.

\[ n(\ce{CaCO3\ left}) =0.1998-0.1875 \]

\[ =0.0123\text{ mol} \]

Convert to mass:

\[ m=0.0123\times100.09 \]

\[ \boxed{m(\ce{CaCO3\ left})=1.23\text{ g}} \]

Step 7: Calculate the total dry residue.

The dry residue contains two substances:

• 1.23 g of unreacted calcium carbonate.
• 5.00 g of inert impurity.

Therefore:

\[ m(\text{residue})=1.23+5.00 \]

\[ \boxed{m(\text{residue})=6.23\text{ g}} \]

The distinction matters. The impurity never reacted, so it cannot be classified as excess reactant.

Step 8: Calculate the theoretical carbon dioxide yield.

The mole ratio between hydrochloric acid and carbon dioxide is 2:1.

\[ n(\ce{CO2})=\frac{0.375}{2} \]

\[ =0.1875\text{ mol} \]

Convert to mass:

\[ m(\ce{CO2})=0.1875\times44.01 \]

\[ \boxed{m(\ce{CO2})=8.25\text{ g}} \]

This question combines limiting reactants with percentage purity. The important extra step is removing the inert material from the initial sample mass before calculating moles.

Question 4

Zinc reacts with hydrochloric acid according to:

\[ \ce{Zn(s) + 2HCl(aq) -> ZnCl2(aq) + H2(g)} \]

A flask initially contains 9.81 g of zinc and 100.0 mL of hydrochloric acid at 2.00 mol L\(^{-1}\).

After the first reaction has gone to completion, an additional 50.0 mL of hydrochloric acid at the same concentration is added.

Assume no zinc or solution is lost between the two stages.

Calculate:

a) The mass of zinc remaining after the first addition of acid.

b) Whether zinc or hydrochloric acid remains after the second addition.

c) The total amount, in moles, of hydrogen gas produced across both stages.

Explain why you must use the remaining zinc amount, rather than the original zinc amount, when analysing the second stage.

Solution 4

After the first addition, 3.27 g of zinc remains. The second addition consumes this remaining zinc exactly, leaving neither reactant in excess. The total hydrogen produced is 0.150 mol.

Step 1: Calculate the initial moles of zinc.

Using \(M(\ce{Zn})=65.4\text{ g mol}^{-1}\):

\[ n(\ce{Zn})=\frac{9.81}{65.4} \]

\[ =0.150\text{ mol} \]

Step 2: Calculate the initial moles of hydrochloric acid.

\[ n(\ce{HCl})=2.00\times0.1000 \]

\[ =0.200\text{ mol} \]

Step 3: Identify the limiting reactant in the first stage.

Divide by the coefficients.

For zinc:

\[ \frac{0.150}{1}=0.150 \]

For hydrochloric acid:

\[ \frac{0.200}{2}=0.100 \]

Hydrochloric acid is limiting.

Step 4: Calculate the zinc consumed.

The equation requires two moles of hydrochloric acid for every mole of zinc.

\[ n(\ce{Zn\ consumed})=\frac{0.200}{2} \]

\[ =0.100\text{ mol} \]

Step 5: Calculate the zinc remaining after the first stage.

\[ n(\ce{Zn\ left})=0.150-0.100 \]

\[ =0.0500\text{ mol} \]

Convert to mass:

\[ m=0.0500\times65.4 \]

\[ \boxed{m(\ce{Zn\ left})=3.27\text{ g}} \]

Step 6: Calculate the additional hydrochloric acid.

The second addition is 50.0 mL at 2.00 mol L\(^{-1}\).

\[ n(\ce{HCl\ added})=2.00\times0.0500 \]

\[ =0.100\text{ mol} \]

Step 7: Determine the final leftover amounts.

There are 0.0500 mol of zinc remaining.

The equation requires two moles of hydrochloric acid for every mole of zinc.

Therefore, the acid required is:

\[ n(\ce{HCl\ required})=0.0500\times2 \]

\[ =0.100\text{ mol} \]

This is exactly the amount added.

Both reactants are completely consumed in the second stage.

\[ \boxed{n(\ce{Zn\ left})=0} \]

\[ \boxed{n(\ce{HCl\ left})=0} \]

Step 8: Calculate the total hydrogen produced.

One mole of zinc produces one mole of hydrogen.

Across both stages, the total zinc consumed is 0.150 mol.

Therefore:

\[ n(\ce{H2})=0.150\text{ mol} \]

The total hydrogen production is:

\[ \boxed{0.150\text{ mol}} \]

The key idea is that the second calculation starts with the amount of zinc remaining after the first reaction. Using the original 0.150 mol again would count the first stage twice.

Question 5

Two identical batches each begin with:

• 5.00 g of pure calcium carbonate.
• 50.0 mL of hydrochloric acid at 1.00 mol L\(^{-1}\).

The reaction is:

\[ \ce{CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l)} \]

A student collects 1.66 g of calcium chloride from each batch. They conclude that both batches must have the same quantities of excess reactants remaining.

However, the two batches were treated differently.

Batch A: The reaction proceeds to completion. However, only 60.0% of the theoretically produced calcium chloride is successfully recovered because some product is lost during transfer and filtration.

Batch B: The reaction is deliberately stopped when it has reached 60.0% of its maximum theoretical reaction extent. All the calcium chloride produced is successfully recovered.

Assume there are no side reactions, and all unrecovered product in Batch A is due to physical recovery losses.

Calculate:

a) The theoretical mass of calcium chloride for a complete reaction.

b) The mass of calcium carbonate remaining in Batch A.

c) The masses of calcium carbonate and hydrochloric acid remaining in Batch B.

d) Explain why the two batches produce the same recovered product mass but have different quantities of excess reactant remaining.

e) Identify one additional measurement that could distinguish the two batches experimentally.

Solution 5

The student’s conclusion is incorrect. Batch A leaves 2.50 g of calcium carbonate and no hydrochloric acid. Batch B leaves 3.50 g of calcium carbonate and approximately 0.729 g of hydrochloric acid.

The two batches have the same recovered product mass, but the reason for their reduced yield is different.

This is an important distinction between percentage yield and reaction conversion.

Step 1: Calculate the initial moles of each reactant.

For calcium carbonate:

\[ n(\ce{CaCO3})=\frac{5.00}{100.09} \]

\[ =0.04996\text{ mol} \]

For hydrochloric acid:

\[ n(\ce{HCl})=1.00\times0.0500 \]

\[ =0.0500\text{ mol} \]

Step 2: Identify the limiting reactant.

Divide by the coefficients.

For calcium carbonate:

\[ \frac{0.04996}{1}=0.04996 \]

For hydrochloric acid:

\[ \frac{0.0500}{2}=0.0250 \]

Hydrochloric acid is limiting.

Therefore, the maximum possible reaction extent is 0.0250 mol.

Step 3: Calculate the theoretical calcium chloride yield.

The equation produces one mole of calcium chloride for every two moles of hydrochloric acid.

\[ n(\ce{CaCl2})=\frac{0.0500}{2} \]

\[ =0.0250\text{ mol} \]

Using \(M(\ce{CaCl2})=110.98\text{ g mol}^{-1}\):

\[ m(\ce{CaCl2})=0.0250\times110.98 \]

\[ \boxed{m(\ce{CaCl2})=2.77\text{ g}} \]

This is the theoretical yield for both batches.

Step 4: Calculate the remaining reactants in Batch A.

Batch A completes the reaction.

Therefore, all 0.0500 mol of hydrochloric acid is consumed.

The calcium carbonate consumed is:

\[ n(\ce{CaCO3\ consumed})=0.0250\text{ mol} \]

The calcium carbonate remaining is:

\[ n(\ce{CaCO3\ left}) =0.04996-0.0250 \]

\[ =0.02496\text{ mol} \]

Convert into mass:

\[ m=0.02496\times100.09 \]

\[ \boxed{m(\ce{CaCO3\ left})=2.50\text{ g}} \]

There is no hydrochloric acid remaining.

The actual recovered calcium chloride is:

\[ m(\ce{CaCl2\ recovered}) =0.600\times2.77 \]

\[ \boxed{m(\ce{CaCl2\ recovered})=1.66\text{ g}} \]

The reaction itself was complete. The missing product was lost during recovery.

Step 5: Calculate the reaction extent for Batch B.

Batch B reaches only 60.0% of the maximum theoretical reaction extent.

Therefore:

\[ n(\text{reaction extent}) =0.600\times0.0250 \]

\[ =0.0150\text{ mol} \]

Only 0.0150 mol of calcium carbonate reacts.

Step 6: Calculate the calcium carbonate remaining in Batch B.

\[ n(\ce{CaCO3\ left}) =0.04996-0.0150 \]

\[ =0.03496\text{ mol} \]

Convert to mass:

\[ m=0.03496\times100.09 \]

\[ \boxed{m(\ce{CaCO3\ left})=3.50\text{ g}} \]

Step 7: Calculate the hydrochloric acid remaining in Batch B.

The reaction requires two moles of hydrochloric acid for every mole of calcium carbonate.

The amount consumed is:

\[ n(\ce{HCl\ consumed})=0.0150\times2 \]

\[ =0.0300\text{ mol} \]

The amount remaining is:

\[ n(\ce{HCl\ left})=0.0500-0.0300 \]

\[ =0.0200\text{ mol} \]

Convert into mass:

\[ m=0.0200\times36.46 \]

\[ \boxed{m(\ce{HCl\ left})=0.729\text{ g}} \]

Step 8: Explain why the recovered product masses are identical.

In Batch B, the amount of calcium chloride actually formed is:

\[ n(\ce{CaCl2})=0.0150\text{ mol} \]

Its mass is:

\[ m=0.0150\times110.98 \]

\[ \boxed{m=1.66\text{ g}} \]

All of this product is recovered.

In Batch A, 2.77 g of calcium chloride is formed, but only 60.0% is collected.

Both batches therefore produce the same recovered mass, even though the amounts of reactants consumed are different.

The crucial distinction is:

• Percentage yield measures how much product is recovered compared with the theoretical maximum.
• Reaction conversion describes how far the chemical reaction has actually proceeded.

A low percentage yield does not automatically mean the reaction was incomplete.

Step 9: Identify an additional experimental measurement.

One useful method would be to titrate the remaining hydrochloric acid in the reaction mixture.

Batch A should contain no hydrochloric acid, whereas Batch B contains 0.0200 mol.

Alternatively, the remaining solid could be filtered, washed, dried, and weighed. The two batches would have different masses of unreacted calcium carbonate.

The important conclusion is that recovered product mass alone cannot establish how much excess reactant remains. You must know whether the product loss occurred during the reaction or during product recovery.

07Where to go next

Once you can calculate the mass of excess reactant, you have the main tools needed for more involved stoichiometry questions.

The next decision is to distinguish between three quantities:

• The limiting reactant, which determines the maximum theoretical reaction extent.
• The excess reactant, which determines what remains after the assumed reaction.
• The actual product yield, which tells you how much product was successfully obtained.

Also remember that this method applies to reactants being consumed in a chemical reaction. If a question instead asks how much of a sparingly soluble ionic solid dissolves at equilibrium, you need a different approach using the solubility product constant, \(K_{sp}\). See Calculate Solubility from Ksp for HSC Chemistry.

For excess-reactant questions, keep returning to one calculation:

\[ \boxed{\frac{\text{initial moles}}{\text{balanced coefficient}}} \]

Find the smallest value, use its mole ratio to calculate what was consumed, and subtract to find what remains. That is the foundation of an accurate excess-reactant calculation.