How to Calculate the Minimum Volume of Acid Needed

To calculate the minimum volume of acid needed to react completely with a given amount of another reactant, first calculate how many moles of acid the balanced equation requires. Then divide that amount by the acid's concentration.

The key step is to divide each reactant's amount by its coefficient in the balanced equation. This tells you which reactant runs out first and how much acid is actually needed. Once you know the required moles of acid, use \(V = n/c\) to calculate its volume.

The distinction is important: having enough acid to start a reaction is not the same as having enough to react completely. If you add too little, some of the other reactant remains. If you add too much, the acid is in excess.

## Work out the acid-to-reactant ratio

Imagine you have 0.0100 mol of sodium hydroxide, \(\ce{NaOH}\), and want to neutralise it using sulfuric acid, \(\ce{H2SO4}\).

You might predict that you need 0.0100 mol of acid because the two amounts should match. But that prediction is wrong.

The balanced equation is:

\[
\ce{H2SO4(aq) + 2NaOH(aq) -> Na2SO4(aq) + 2H2O(l)}
\]

One mole of sulfuric acid reacts with two moles of sodium hydroxide. Therefore, 0.0100 mol of \(\ce{NaOH}\) needs only 0.00500 mol of \(\ce{H2SO4}\).

The coefficients determine the required ratio. They are not just numbers added to balance an equation.

[IMAGE: An annotated reaction illustration showing one mole of sulfuric acid reacting with two moles of sodium hydroxide to form one mole of sodium sulfate and two moles of water. Clearly distinguish the 1:2:1:2 mole ratio and show the acid and base as separate reactant quantities.]

### Why divide the amount by the coefficient?

Consider a general reaction:

\[
\ce{aA + bB -> products}
\]

The coefficients tell you how many moles of each reactant are needed for one complete reaction portion.

For example, in the sulfuric acid reaction:

- One reaction portion requires 1 mol of \(\ce{H2SO4}\).
- One reaction portion requires 2 mol of \(\ce{NaOH}\).

To find how many complete reaction portions each reactant can support, divide its amount by its coefficient.

\[
\text{Reaction portions}=
\frac{\text{amount in mol}}{\text{coefficient}}
\]

The reactant with the smaller result is the limiting reactant. It runs out first.

For a reaction between a known amount of another reactant and an acid, the minimum acid requirement occurs when both reactants are present in exactly the balanced equation ratio.

Mathematically:

\[
\frac{n_{\text{target}}}{\nu_{\text{target}}}
=
\frac{n_{\text{acid}}}{\nu_{\text{acid}}}
\]

Rearranging gives:

\[
n_{\text{acid, required}}
=
n_{\text{target}}
\times
\frac{\nu_{\text{acid}}}{\nu_{\text{target}}}
\]

Here:

- \(n_{\text{target}}\) is the amount of the reactant you want to consume completely, in mol.
- \(n_{\text{acid, required}}\) is the required amount of acid, in mol.
- \(\nu_{\text{target}}\) is the target reactant's balanced equation coefficient.
- \(\nu_{\text{acid}}\) is the acid's balanced equation coefficient.

This gives you the amount of acid needed. You can now convert it into a volume.

### The minimum volume formula

The relationship between amount, concentration, and volume is:

\[
n=cV
\]

Rearranging:

\[
V=\frac{n}{c}
\]

Combining this with the mole ratio gives the formula you need:

\[
\boxed{
V_{\min}=
\frac{n_{\text{target}}}{c_{\text{acid}}}
\times
\frac{\nu_{\text{acid}}}{\nu_{\text{target}}}
}
\]

The symbols are:

- \(V_{\min}\): minimum acid volume, in L.
- \(n_{\text{target}}\): initial amount of the other reactant, in mol.
- \(c_{\text{acid}}\): acid concentration, in mol L\(^{-1}\).
- \(\nu_{\text{acid}}\): acid coefficient.
- \(\nu_{\text{target}}\): coefficient of the other reactant.

**Important:** Use litres when calculating volume with \(n=cV\). Convert your final answer into millilitres if needed.

If the target reactant is provided as a mass, first calculate its amount using:

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

where \(m\) is mass in grams and \(M\) is molar mass in g mol\(^{-1}\).

If you want a more detailed explanation of comparing reactants using their mole amounts and coefficients, see [How to Compare Reactant Amounts and Find the Limiting Reactant](https://tutorgum.com/learn/resources/hsc-chemistry-how-to-compare-reactant-quantities-when-equation-coefficients-are-different/).

## Worked example 1: Calculate the minimum volume of sulfuric acid

A student has 25.0 mL of 0.400 mol L\(^{-1}\) sodium hydroxide solution. They want to neutralise it completely using 0.250 mol L\(^{-1}\) sulfuric acid.

Calculate:

a) The minimum volume of sulfuric acid required.

b) The theoretical mass of sodium sulfate produced.

Assume complete neutralisation to sodium sulfate.

[IMAGE: A conical flask containing 25.0 mL of sodium hydroxide solution beneath a burette containing sulfuric acid. Show the acid being added dropwise while the flask is swirled. Label both concentrations and the initial base volume.]

**Step 1**: Write the balanced chemical equation.

\[
\ce{H2SO4 + 2NaOH -> Na2SO4 + 2H2O}
\]

The mole ratio is:

\[
\ce{H2SO4 : NaOH}=1:2
\]

**Step 2**: Calculate the initial amount of sodium hydroxide.

We know:

- \(c=0.400\ \mathrm{mol\,L^{-1}}\)
- \(V=25.0\ \mathrm{mL}=0.0250\ \mathrm L\)

Using \(n=cV\):

\[
\begin{aligned}
n(\ce{NaOH})&=0.400\times0.0250\\
&=0.0100\ \mathrm{mol}
\end{aligned}
\]

**Step 3**: Calculate the amount of sulfuric acid required.

The equation requires one mole of sulfuric acid for every two moles of sodium hydroxide.

\[
\begin{aligned}
n(\ce{H2SO4})
&=0.0100\times\frac{1}{2}\\
&=0.00500\ \mathrm{mol}
\end{aligned}
\]

**Step 4**: Convert the required acid amount into volume.

The acid concentration is 0.250 mol L\(^{-1}\).

\[
\begin{aligned}
V&=\frac{n}{c}\\
&=\frac{0.00500}{0.250}\\
&=0.0200\ \mathrm L
\end{aligned}
\]

Convert to millilitres:

\[
V=0.0200\times1000
\]

\[
\boxed{V_{\min}=20.0\ \mathrm{mL}}
\]

This is the minimum theoretical volume of sulfuric acid needed to neutralise all the sodium hydroxide.

**Step 5**: Calculate the theoretical yield of sodium sulfate.

The equation shows a 1:1 mole ratio between sulfuric acid and sodium sulfate.

Therefore:

\[
n(\ce{Na2SO4})=0.00500\ \mathrm{mol}
\]

The molar mass of sodium sulfate is:

\[
M(\ce{Na2SO4})=142.0\ \mathrm{g\,mol^{-1}}
\]

Using \(m=nM\):

\[
\begin{aligned}
m&=0.00500\times142.0\\
&=0.710\ \mathrm g
\end{aligned}
\]

The theoretical yield is 0.710 g of sodium sulfate.

### What if you add less than the minimum?

Suppose the student accidentally adds only 15.0 mL of sulfuric acid.

The acid amount is now:

\[
n(\ce{H2SO4})=0.250\times0.0150
\]

\[
=0.00375\ \mathrm{mol}
\]

Compare the reactants by dividing each amount by its coefficient.

| Reactant | Initial amount (mol) | Coefficient | Amount ÷ coefficient |
|---|---:|---:|---:|
| \(\ce{H2SO4}\) | 0.00375 | 1 | 0.00375 |
| \(\ce{NaOH}\) | 0.0100 | 2 | 0.00500 |

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

The reaction stops when the acid runs out.

The amount of sodium hydroxide consumed is:

\[
n(\ce{NaOH\ consumed})
=0.00375\times2
\]

\[
=0.00750\ \mathrm{mol}
\]

The amount remaining is:

\[
\begin{aligned}
n(\ce{NaOH\ left})
&=0.0100-0.00750\\
&=0.00250\ \mathrm{mol}
\end{aligned}
\]

The theoretical amount of sodium sulfate produced is also 0.00375 mol, corresponding to 0.533 g.

This is why you cannot calculate product yield from the initial amount of sodium hydroxide alone. You must account for the acid available.

## Worked example 2: Calculate acid volume, leftover solid, and percentage yield

A student reacts 1.50 g of pure calcium carbonate, \(\ce{CaCO3}\), with 40.0 mL of 0.600 mol L\(^{-1}\) hydrochloric acid.

The reaction produces carbon dioxide.

The student collects 0.400 g of carbon dioxide.

Calculate:

a) The minimum volume of hydrochloric acid needed to react completely with the calcium carbonate.

b) The mass of calcium carbonate left over when only 40.0 mL of acid is added.

c) The theoretical yield of carbon dioxide.

d) The percentage yield of carbon dioxide.

Use:

- \(M(\ce{CaCO3})=100.1\ \mathrm{g\,mol^{-1}}\)
- \(M(\ce{CO2})=44.0\ \mathrm{g\,mol^{-1}}\)

[IMAGE: A conical flask containing visible calcium carbonate marble chips and hydrochloric acid, connected through a delivery tube to a graduated gas syringe collecting carbon dioxide. Label the initial solid mass, acid concentration, acid volume, and collected gas.]

**Step 1**: Write the balanced chemical equation.

\[
\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}
\]

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

**Step 2**: Calculate the initial amount of calcium carbonate.

Use:

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

Substitute:

\[
\begin{aligned}
n(\ce{CaCO3})
&=\frac{1.50}{100.1}\\
&=0.01499\ \mathrm{mol}
\end{aligned}
\]

**Step 3**: Calculate the minimum amount of hydrochloric acid.

The mole ratio is 1:2.

\[
\begin{aligned}
n(\ce{HCl, required})
&=0.01499\times\frac{2}{1}\\
&=0.02997\ \mathrm{mol}
\end{aligned}
\]

**Step 4**: Convert the required acid amount into volume.

The hydrochloric acid concentration is 0.600 mol L\(^{-1}\).

\[
\begin{aligned}
V_{\min}
&=\frac{0.02997}{0.600}\\
&=0.04995\ \mathrm L\\
&=49.95\ \mathrm{mL}
\end{aligned}
\]

To three significant figures:

\[
\boxed{V_{\min}=50.0\ \mathrm{mL}}
\]

The student needs approximately 50.0 mL of hydrochloric acid to react completely with the original calcium carbonate.

However, they only used 40.0 mL.

**Step 5**: Determine which reactant is limiting.

Calculate the actual amount of hydrochloric acid supplied:

\[
\begin{aligned}
n(\ce{HCl})
&=0.600\times0.0400\\
&=0.0240\ \mathrm{mol}
\end{aligned}
\]

Compare the two reactants:

| Reactant | Initial amount (mol) | Coefficient | Amount ÷ coefficient |
|---|---:|---:|---:|
| \(\ce{CaCO3}\) | 0.01499 | 1 | 0.01499 |
| \(\ce{HCl}\) | 0.0240 | 2 | 0.0120 |

Hydrochloric acid has the smaller result.

It is the limiting reactant, meaning some calcium carbonate will remain unreacted.

**Step 6**: Calculate the mass of calcium carbonate left over.

The amount of calcium carbonate that reacts is:

\[
n(\ce{CaCO3\ consumed})
=\frac{0.0240}{2}
\]

\[
=0.0120\ \mathrm{mol}
\]

Subtract this from the initial amount:

\[
\begin{aligned}
n(\ce{CaCO3\ left})
&=0.01499-0.0120\\
&=0.00299\ \mathrm{mol}
\end{aligned}
\]

Convert the remaining amount into mass:

\[
\begin{aligned}
m&=nM\\
&=0.00299\times100.1\\
&=0.299\ \mathrm g
\end{aligned}
\]

Approximately 0.299 g of calcium carbonate remains.

**Step 7**: Calculate the theoretical yield of carbon dioxide.

The equation gives a 2:1 mole ratio between hydrochloric acid and carbon dioxide.

Therefore:

\[
n(\ce{CO2})=\frac{0.0240}{2}
\]

\[
=0.0120\ \mathrm{mol}
\]

Convert this into mass:

\[
\begin{aligned}
m(\ce{CO2})
&=nM\\
&=0.0120\times44.0\\
&=0.528\ \mathrm g
\end{aligned}
\]

The theoretical yield is 0.528 g of carbon dioxide.

Notice that we calculated this using the limiting reactant, hydrochloric acid, rather than the initial amount of calcium carbonate.

**Step 8**: Calculate the percentage yield.

The student actually collected 0.400 g of carbon dioxide.

Use:

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

Substitute:

\[
\begin{aligned}
\%\text{ yield}
&=\frac{0.400}{0.528}\times100\\
&=75.8\%
\end{aligned}
\]

The percentage yield is 75.8%.

This is the percentage of the theoretically possible carbon dioxide that was actually collected.

### What if the acid is in excess?

Suppose the student instead uses 60.0 mL of 0.600 mol L\(^{-1}\) hydrochloric acid.

The acid amount is:

\[
n(\ce{HCl})=0.600\times0.0600
\]

\[
=0.0360\ \mathrm{mol}
\]

The calcium carbonate requires only 0.02997 mol of acid.

Therefore, calcium carbonate is now the limiting reactant.

The excess acid remaining is:

\[
\begin{aligned}
n(\ce{HCl\ left})
&=0.0360-0.02997\\
&=0.00603\ \mathrm{mol}
\end{aligned}
\]

All the calcium carbonate reacts, and hydrochloric acid remains in excess.

[IMAGE: Three side-by-side reaction vessels containing the same initial mass of calcium carbonate, with 40.0 mL, approximately 50.0 mL, and 60.0 mL of 0.600 M hydrochloric acid respectively. Show unreacted solid in the acid-deficient vessel, complete consumption at the stoichiometric amount, and excess acid in the final vessel. Label each condition clearly.]

The important distinction is:

- Too little acid: some of the target reactant remains.
- Exactly the required amount: both reactants are completely consumed.
- Too much acid: the target reactant is consumed, but some acid remains.

## A reliable method for any minimum acid volume question

Use this sequence when tackling an HSC Chemistry calculation.

| Step | What to do | Why it matters |
|---|---|---|
| 1 | Write the balanced equation. | Establishes the correct mole ratio. |
| 2 | Convert the known reactant quantity into moles. | Stoichiometric ratios compare moles, not grams or millilitres. |
| 3 | Divide the amount by its coefficient. | Establishes how much reaction each reactant can support. |
| 4 | Calculate the required acid moles. | Gives the exact stoichiometric acid requirement. |
| 5 | Use \(V=n/c\). | Converts the required acid amount into volume. |
| 6 | Check the actual acid supplied, if given. | Determines whether the acid is limiting or in excess. |
| 7 | Calculate leftovers and product yield. | Ensures the final result reflects what can actually react. |

### Three common mistakes to avoid

**1. Comparing raw mole amounts**

Suppose you have 0.0200 mol of magnesium and 0.0300 mol of hydrochloric acid.

You might assume magnesium is limiting because it has fewer moles.

But the equation is:

\[
\ce{Mg + 2HCl -> MgCl2 + H2}
\]

The acid needs to be present in twice the amount of magnesium. Always account for the coefficients.

**2. Forgetting to convert millilitres into litres**

If you use \(n=cV\), the volume must be in litres.

For example:

\[
50.0\ \mathrm{mL}=0.0500\ \mathrm L
\]

Using 50.0 instead of 0.0500 would make your calculated amount 1000 times too large.

**3. Calculating theoretical yield from the wrong reactant**

If acid is limiting, you cannot calculate product yield as though all the original solid reacted.

Use the limiting reactant to calculate the theoretical product amount. Otherwise, your theoretical yield will be too high, and your percentage yield will be misleading.

### Does the type of acid change the calculation?

Yes, the balanced equation can change the required mole ratio.

For example, sulfuric acid is diprotic, meaning each formula unit contains two ionisable hydrogen atoms.

For complete neutralisation with sodium hydroxide:

\[
\ce{H2SO4 + 2NaOH -> Na2SO4 + 2H2O}
\]

One mole of sulfuric acid reacts with two moles of sodium hydroxide.

However, if the reaction produces sodium hydrogen sulfate instead:

\[
\ce{H2SO4 + NaOH -> NaHSO4 + H2O}
\]

The ratio is now 1:1.

Don't automatically multiply by the number of hydrogen atoms in an acid. Use the balanced equation and the products specified in the question.

### Assumptions behind the minimum volume

The calculated minimum volume is a theoretical stoichiometric quantity. It assumes:

- The chemical equation is balanced correctly.
- The reactants have the stated purity and concentration.
- The reaction proceeds to completion.
- No competing reactions consume additional acid.
- The acid volume and concentration are measured accurately.

In a practical experiment, a small excess of acid may sometimes be used to ensure complete reaction. That does not change the theoretical minimum.

## Questions and solutions

### Question 1

A student has 0.0150 mol of sodium hydroxide and wants to neutralise it completely using 0.200 mol L\(^{-1}\) hydrochloric acid.

The balanced equation is:

\[
\ce{HCl + NaOH -> NaCl + H2O}
\]

Calculate:

a) The minimum volume of hydrochloric acid required.

b) The theoretical mass of sodium chloride produced.

Use \(M(\ce{NaCl})=58.44\ \mathrm{g\,mol^{-1}}\).

### Solution 1

The minimum volume is 75.0 mL, and the theoretical yield is 0.877 g of sodium chloride.

**Step 1**: Identify the mole ratio.

The equation gives a 1:1 ratio between hydrochloric acid and sodium hydroxide.

Therefore:

\[
n(\ce{HCl})=0.0150\ \mathrm{mol}
\]

**Step 2**: Calculate the minimum acid volume.

\[
\begin{aligned}
V&=\frac{n}{c}\\
&=\frac{0.0150}{0.200}\\
&=0.0750\ \mathrm L
\end{aligned}
\]

Convert to millilitres:

\[
V=0.0750\times1000
\]

\[
\boxed{V_{\min}=75.0\ \mathrm{mL}}
\]

**Step 3**: Calculate the theoretical sodium chloride yield.

The mole ratio between hydrochloric acid and sodium chloride is also 1:1.

\[
n(\ce{NaCl})=0.0150\ \mathrm{mol}
\]

Using \(m=nM\):

\[
\begin{aligned}
m&=0.0150\times58.44\\
&=0.8766\ \mathrm g
\end{aligned}
\]

\[
\boxed{m(\ce{NaCl})=0.877\ \mathrm g}
\]

Both reactants are completely consumed at the calculated minimum volume, assuming an ideal reaction.

### Question 2

A student places 1.20 g of magnesium ribbon into a flask containing 150.0 mL of 0.500 mol L\(^{-1}\) hydrochloric acid.

The balanced equation is:

\[
\ce{Mg + 2HCl -> MgCl2 + H2}
\]

Calculate:

a) The minimum volume of hydrochloric acid required to react completely with the magnesium.

b) Whether the 150.0 mL supplied is sufficient.

c) The mass of magnesium remaining after the reaction.

d) The theoretical volume of hydrogen produced at 25°C and 100 kPa.

Use:

- \(M(\ce{Mg})=24.3\ \mathrm{g\,mol^{-1}}\)
- Molar gas volume at 25°C and 100 kPa = 24.8 L mol\(^{-1}\).

[IMAGE: Magnesium ribbon suspended in a conical flask containing hydrochloric acid, connected by a sealed delivery tube to a graduated gas syringe. Label the initial magnesium mass, acid concentration, acid volume, and hydrogen collection apparatus.]

### Solution 2

The minimum acid volume is 198 mL. The 150.0 mL supplied is insufficient, leaving approximately 0.289 g of magnesium unreacted. The theoretical hydrogen volume is 0.930 L.

**Step 1**: Calculate the initial amount of magnesium.

\[
\begin{aligned}
n(\ce{Mg})
&=\frac{m}{M}\\
&=\frac{1.20}{24.3}\\
&=0.0494\ \mathrm{mol}
\end{aligned}
\]

**Step 2**: Calculate the hydrochloric acid required.

The equation requires two moles of hydrochloric acid per mole of magnesium.

\[
\begin{aligned}
n(\ce{HCl, required})
&=0.0494\times2\\
&=0.0988\ \mathrm{mol}
\end{aligned}
\]

**Step 3**: Calculate the minimum volume.

\[
\begin{aligned}
V_{\min}
&=\frac{0.0988}{0.500}\\
&=0.1976\ \mathrm L\\
&=197.6\ \mathrm{mL}
\end{aligned}
\]

To three significant figures:

\[
\boxed{V_{\min}=198\ \mathrm{mL}}
\]

**Step 4**: Check whether the supplied acid is sufficient.

The actual amount of hydrochloric acid is:

\[
\begin{aligned}
n(\ce{HCl})
&=0.500\times0.1500\\
&=0.0750\ \mathrm{mol}
\end{aligned}
\]

Divide each amount by its coefficient:

\[
\frac{n(\ce{Mg})}{1}=0.0494
\]

\[
\frac{n(\ce{HCl})}{2}=0.0375
\]

Hydrochloric acid has the smaller result.

It is the limiting reactant, so the magnesium cannot react completely.

**Step 5**: Calculate the remaining magnesium.

The amount of magnesium consumed is:

\[
n(\ce{Mg\ consumed})=\frac{0.0750}{2}
\]

\[
=0.0375\ \mathrm{mol}
\]

The amount remaining is:

\[
\begin{aligned}
n(\ce{Mg\ left})
&=0.0494-0.0375\\
&=0.0119\ \mathrm{mol}
\end{aligned}
\]

Convert to mass:

\[
\begin{aligned}
m&=0.0119\times24.3\\
&=0.289\ \mathrm g
\end{aligned}
\]

Approximately 0.289 g of magnesium remains.

**Step 6**: Calculate the theoretical hydrogen volume.

The balanced equation gives a 2:1 ratio between hydrochloric acid and hydrogen.

\[
n(\ce{H2})=\frac{0.0750}{2}
\]

\[
=0.0375\ \mathrm{mol}
\]

Using the supplied molar gas volume:

\[
\begin{aligned}
V(\ce{H2})
&=nV_m\\
&=0.0375\times24.8\\
&=0.930\ \mathrm L
\end{aligned}
\]

\[
\boxed{V(\ce{H2})=0.930\ \mathrm L}
\]

The hydrogen yield is limited by the hydrochloric acid, not the original magnesium mass.

### Question 3

A 2.00 g impure sample of magnesium oxide contains 85.0% magnesium oxide by mass. It reacts with 500 mL of 0.200 mol L\(^{-1}\) hydrochloric acid.

The balanced equation is:

\[
\ce{MgO + 2HCl -> MgCl2 + H2O}
\]

Calculate:

a) The minimum volume of hydrochloric acid needed to react completely with the magnesium oxide.

b) The amount of hydrochloric acid remaining after the reaction.

c) The theoretical mass of magnesium chloride produced.

Use:

- \(M(\ce{MgO})=40.3\ \mathrm{g\,mol^{-1}}\)
- \(M(\ce{MgCl2})=95.2\ \mathrm{g\,mol^{-1}}\)

Assume the impurities do not react with hydrochloric acid.

[IMAGE: A laboratory weighing boat containing a labelled 2.00 g impure magnesium oxide sample beside a flask of hydrochloric acid. Show the sample being transferred into the flask, with the 85.0% purity and acid concentration identified.]

### Solution 3

The minimum volume is 422 mL. The 500 mL supplied is in excess, leaving 0.0156 mol of hydrochloric acid. The theoretical yield is 4.02 g of magnesium chloride.

**Step 1**: Calculate the actual mass of magnesium oxide.

Only 85.0% of the sample is magnesium oxide.

\[
\begin{aligned}
m(\ce{MgO})
&=2.00\times0.850\\
&=1.70\ \mathrm g
\end{aligned}
\]

**Step 2**: Convert the pure magnesium oxide mass into moles.

\[
\begin{aligned}
n(\ce{MgO})
&=\frac{1.70}{40.3}\\
&=0.0422\ \mathrm{mol}
\end{aligned}
\]

**Step 3**: Calculate the hydrochloric acid requirement.

The equation gives a 1:2 ratio.

\[
\begin{aligned}
n(\ce{HCl, required})
&=0.0422\times2\\
&=0.0844\ \mathrm{mol}
\end{aligned}
\]

**Step 4**: Calculate the minimum acid volume.

\[
\begin{aligned}
V_{\min}
&=\frac{0.0844}{0.200}\\
&=0.422\ \mathrm L
\end{aligned}
\]

Therefore:

\[
\boxed{V_{\min}=422\ \mathrm{mL}}
\]

**Step 5**: Calculate the amount of acid initially supplied.

Convert 500 mL into litres:

\[
500\ \mathrm{mL}=0.500\ \mathrm L
\]

Then:

\[
\begin{aligned}
n(\ce{HCl, initial})
&=0.200\times0.500\\
&=0.100\ \mathrm{mol}
\end{aligned}
\]

The supplied acid amount is greater than the required amount.

Hydrochloric acid is in excess, and all the magnesium oxide reacts.

The remaining acid is:

\[
\begin{aligned}
n(\ce{HCl, left})
&=0.100-0.0844\\
&=0.0156\ \mathrm{mol}
\end{aligned}
\]

**Step 6**: Calculate the theoretical yield of magnesium chloride.

The equation gives a 1:1 ratio between magnesium oxide and magnesium chloride.

\[
n(\ce{MgCl2})=0.0422\ \mathrm{mol}
\]

Using the molar mass:

\[
\begin{aligned}
m(\ce{MgCl2})
&=0.0422\times95.2\\
&=4.02\ \mathrm g
\end{aligned}
\]

\[
\boxed{m(\ce{MgCl2})=4.02\ \mathrm g}
\]

The theoretical yield is based on the pure magnesium oxide, not the entire impure sample.

### Question 4

A laboratory has 100.0 mL of 0.600 mol L\(^{-1}\) hydrochloric acid stock solution. The acid is diluted with water to a final volume of 250.0 mL.

A student takes 40.0 mL of the diluted acid and adds it to 0.500 g of magnesium.

The reaction is:

\[
\ce{Mg + 2HCl -> MgCl2 + H2}
\]

Calculate:

a) The minimum volume of diluted hydrochloric acid needed to react completely with the magnesium.

b) The mass of magnesium remaining after using 40.0 mL of diluted acid.

c) The theoretical volume of hydrogen produced at 25°C and 100 kPa.

d) The equivalent volume of the original stock acid required to provide the minimum amount of hydrochloric acid.

Use:

- \(M(\ce{Mg})=24.3\ \mathrm{g\,mol^{-1}}\)
- Molar gas volume at 25°C and 100 kPa = 24.8 L mol\(^{-1}\).

[IMAGE: A volumetric flask containing diluted hydrochloric acid, with a pipette transferring a measured 40.0 mL aliquot into a reaction flask containing magnesium ribbon. Show the original stock acid bottle, dilution flask, reaction flask, and a connected hydrogen gas syringe in their correct spatial order.]

### Solution 4

The minimum volume of diluted acid is 171 mL. Using only 40.0 mL leaves approximately 0.383 g of magnesium. The theoretical hydrogen volume is 0.119 L, and the required stock acid volume is 68.6 mL.

**Step 1**: Calculate the concentration after dilution.

Use the dilution relationship:

\[
c_1V_1=c_2V_2
\]

Where:

- \(c_1=0.600\ \mathrm{mol\,L^{-1}}\)
- \(V_1=0.1000\ \mathrm L\)
- \(V_2=0.2500\ \mathrm L\)
- \(c_2\) is the new concentration.

Rearrange:

\[
\begin{aligned}
c_2
&=\frac{c_1V_1}{V_2}\\
&=\frac{0.600\times0.1000}{0.2500}\\
&=0.240\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]

The diluted acid concentration is 0.240 mol L\(^{-1}\).

**Step 2**: Calculate the amount of magnesium.

\[
\begin{aligned}
n(\ce{Mg})
&=\frac{0.500}{24.3}\\
&=0.0206\ \mathrm{mol}
\end{aligned}
\]

**Step 3**: Calculate the required acid amount.

The equation requires two moles of hydrochloric acid per mole of magnesium.

\[
\begin{aligned}
n(\ce{HCl, required})
&=0.0206\times2\\
&=0.0412\ \mathrm{mol}
\end{aligned}
\]

**Step 4**: Calculate the minimum volume of diluted acid.

\[
\begin{aligned}
V_{\min}
&=\frac{0.0412}{0.240}\\
&=0.171\ \mathrm L
\end{aligned}
\]

Therefore:

\[
\boxed{V_{\min}=171\ \mathrm{mL}}
\]

**Step 5**: Determine what happens when only 40.0 mL is used.

Calculate the actual amount of acid:

\[
\begin{aligned}
n(\ce{HCl})
&=0.240\times0.0400\\
&=0.00960\ \mathrm{mol}
\end{aligned}
\]

Divide each amount by its coefficient:

\[
\frac{n(\ce{Mg})}{1}=0.0206
\]

\[
\frac{n(\ce{HCl})}{2}=0.00480
\]

Hydrochloric acid is limiting.

The amount of magnesium consumed is:

\[
n(\ce{Mg\ consumed})
=\frac{0.00960}{2}
\]

\[
=0.00480\ \mathrm{mol}
\]

The amount remaining is:

\[
\begin{aligned}
n(\ce{Mg\ left})
&=0.0206-0.00480\\
&=0.0158\ \mathrm{mol}
\end{aligned}
\]

Convert to mass:

\[
\begin{aligned}
m(\ce{Mg\ left})
&=0.0158\times24.3\\
&=0.383\ \mathrm g
\end{aligned}
\]

**Step 6**: Calculate the theoretical hydrogen volume.

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

\[
n(\ce{H2})=\frac{0.00960}{2}
\]

\[
=0.00480\ \mathrm{mol}
\]

Using the molar gas volume:

\[
\begin{aligned}
V(\ce{H2})
&=0.00480\times24.8\\
&=0.119\ \mathrm L
\end{aligned}
\]

**Step 7**: Calculate the equivalent stock acid volume.

The minimum amount of hydrochloric acid required was 0.0412 mol.

Using the original stock concentration:

\[
\begin{aligned}
V_{\text{stock}}
&=\frac{0.0412}{0.600}\\
&=0.0686\ \mathrm L\\
&=68.6\ \mathrm{mL}
\end{aligned}
\]

Therefore, 68.6 mL of the original stock solution contains the same number of moles of hydrochloric acid as 171 mL of the diluted solution.

Dilution changes concentration and volume, but it does not change the number of moles of acid in the original solution.

### Question 5

A chemist has a 1.50 g sample containing a mixture of anhydrous sodium carbonate, \(\ce{Na2CO3}\), and sodium hydrogen carbonate, \(\ce{NaHCO3}\).

In a preliminary experiment, the sample reacts completely with excess hydrochloric acid and produces 0.0150 mol of carbon dioxide.

The two reactions are:

\[
\ce{Na2CO3 + 2HCl -> 2NaCl + H2O + CO2}
\]

\[
\ce{NaHCO3 + HCl -> NaCl + H2O + CO2}
\]

The chemist now wants to react a fresh, identical sample completely using 0.400 mol L\(^{-1}\) hydrochloric acid.

Calculate:

a) The initial amounts, in moles, of sodium carbonate and sodium hydrogen carbonate in the sample.

b) The minimum volume of hydrochloric acid required.

c) Whether 60.0 mL of the acid would be sufficient.

d) The theoretical mass of sodium chloride produced.

e) The percentage yield if 1.30 g of dry sodium chloride is recovered when the calculated minimum volume is used.

Use:

- \(M(\ce{Na2CO3})=106.0\ \mathrm{g\,mol^{-1}}\)
- \(M(\ce{NaHCO3})=84.0\ \mathrm{g\,mol^{-1}}\)
- \(M(\ce{NaCl})=58.44\ \mathrm{g\,mol^{-1}}\)

Assume the sample contains only these two salts, all the carbon dioxide is measured in the preliminary experiment, and the recovered sodium chloride is completely dry.

[IMAGE: A conical flask containing a mixed solid sample of sodium carbonate and sodium hydrogen carbonate, connected to a gas syringe through a delivery tube. Show hydrochloric acid being added to the flask and carbon dioxide entering the syringe. Label the total sample mass and measured carbon dioxide amount, but do not indicate the unknown proportions of the two salts.]

### Solution 5

The minimum acid volume is 64.8 mL. The 60.0 mL supply is insufficient, the theoretical sodium chloride yield is 1.51 g, and the percentage yield is 85.9%.

This problem has an important twist: the two salts produce the same amount of carbon dioxide per mole, but they do not consume the same amount of hydrochloric acid.

You must determine the mixture's composition before calculating its acid requirement.

**Step 1**: Define the unknown amounts.

Let:

- \(x\) = amount of sodium carbonate, in mol.
- \(y\) = amount of sodium hydrogen carbonate, in mol.

Both salts produce one mole of carbon dioxide per mole of salt.

Therefore:

\[
x+y=0.0150
\]

This is our first equation.

**Step 2**: Use the total sample mass to create a second equation.

The sample contains 1.50 g of both salts combined.

Using \(m=nM\):

\[
106.0x+84.0y=1.50
\]

We now have two simultaneous equations:

\[
x+y=0.0150
\]

\[
106.0x+84.0y=1.50
\]

**Step 3**: Solve for the amount of each salt.

Rearrange the first equation:

\[
y=0.0150-x
\]

Substitute into the mass equation:

\[
106.0x+84.0(0.0150-x)=1.50
\]

Expand:

\[
106.0x+1.260-84.0x=1.50
\]

Collect like terms:

\[
22.0x=0.240
\]

Therefore:

\[
x=0.0109\ \mathrm{mol}
\]

Substitute this into the first equation:

\[
\begin{aligned}
y&=0.0150-0.0109\\
&=0.00409\ \mathrm{mol}
\end{aligned}
\]

The sample contains:

- 0.0109 mol of \(\ce{Na2CO3}\).
- 0.00409 mol of \(\ce{NaHCO3}\).

**Step 4**: Calculate the total hydrochloric acid requirement.

This is where the two different reaction equations matter.

For sodium carbonate:

\[
n(\ce{HCl})=2x
\]

\[
=2\times0.0109
\]

\[
=0.0218\ \mathrm{mol}
\]

For sodium hydrogen carbonate:

\[
n(\ce{HCl})=y
\]

\[
=0.00409\ \mathrm{mol}
\]

Add the acid requirements:

\[
\begin{aligned}
n(\ce{HCl, required})
&=0.0218+0.00409\\
&=0.0259\ \mathrm{mol}
\end{aligned}
\]

The total amount of hydrochloric acid required is 0.0259 mol.

**Step 5**: Calculate the minimum acid volume.

The hydrochloric acid concentration is 0.400 mol L\(^{-1}\).

\[
\begin{aligned}
V_{\min}
&=\frac{0.0259}{0.400}\\
&=0.0648\ \mathrm L
\end{aligned}
\]

Convert into millilitres:

\[
\boxed{V_{\min}=64.8\ \mathrm{mL}}
\]

**Step 6**: Check whether 60.0 mL is sufficient.

Calculate the amount of acid in 60.0 mL:

\[
\begin{aligned}
n(\ce{HCl})
&=0.400\times0.0600\\
&=0.0240\ \mathrm{mol}
\end{aligned}
\]

The mixture needs 0.0259 mol.

Therefore, 60.0 mL is insufficient.

The acid shortfall is:

\[
\begin{aligned}
n(\ce{HCl, shortfall})
&=0.0259-0.0240\\
&=0.00191\ \mathrm{mol}
\end{aligned}
\]

**Step 7**: Calculate the theoretical yield of sodium chloride.

Look at both balanced equations.

One mole of sodium carbonate produces two moles of sodium chloride.

One mole of sodium hydrogen carbonate produces one mole of sodium chloride.

Therefore:

\[
n(\ce{NaCl})=2x+y
\]

Substitute:

\[
\begin{aligned}
n(\ce{NaCl})
&=2(0.0109)+0.00409\\
&=0.0259\ \mathrm{mol}
\end{aligned}
\]

Convert into mass:

\[
\begin{aligned}
m(\ce{NaCl})
&=0.0259\times58.44\\
&=1.51\ \mathrm g
\end{aligned}
\]

The theoretical yield is 1.51 g of sodium chloride.

**Step 8**: Calculate the percentage yield.

The actual mass recovered is 1.30 g.

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

\[
\begin{aligned}
\%\text{ yield}
&=\frac{1.30}{1.51}\times100\\
&=85.9\%
\end{aligned}
\]

\[
\boxed{\%\text{ yield}=85.9\%}
\]

### Why the obvious shortcut fails

A student might notice that 0.0150 mol of carbon dioxide was produced and assume that the mixture contains 0.0150 mol of sodium carbonate.

They would then calculate:

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

\[
=0.0300\ \mathrm{mol}
\]

This would give an acid volume of:

\[
V=\frac{0.0300}{0.400}
\]

\[
=0.0750\ \mathrm L
\]

That is 75.0 mL, not the correct 64.8 mL.

The problem is the assumption that every mole of carbon dioxide came from sodium carbonate.

Sodium hydrogen carbonate also produces one mole of carbon dioxide per mole, but it requires only one mole of hydrochloric acid rather than two.

The carbon dioxide measurement tells you the total number of moles of the two salts. The mass measurement lets you work out how much of each salt is present. Only then can you calculate the correct acid requirement.

This is why the balanced equation must guide every step of a stoichiometric calculation.

---

Once you have calculated the minimum acid volume, the next decision is usually what the question wants you to determine from the reaction: the mass of excess reactant remaining, the theoretical product yield, or the volume of gas produced.

For carbonate reactions, the next useful calculation is often converting the carbon dioxide amount into a gas volume. See [How to Calculate CO₂ Volume from a Carbonate Reaction](https://tutorgum.com/learn/resources/one-reaction-three-quantities-linking-reacting-mass-solution-concentration-and-gas-volume/).