How to Find the Limiting Reagent When Mixing Two Solutions

Mix 40.0 mL of 0.150 mol L⁻¹ calcium chloride solution with 25.0 mL of 0.100 mol L⁻¹ sodium phosphate solution. The calcium chloride is more concentrated, and you have a larger volume of it. Does that mean calcium chloride must be in excess?

No. To find the limiting reagent when two solutions are mixed, calculate the moles of each reactant using \(n=cV\), divide each amount by its coefficient in the balanced equation, and identify the smallest result. That reactant is the limiting reagent. You can then calculate the amount of product formed and how much of the excess reactant remains.

All numerical data in this article are illustrative.

Before calculating, make a prediction. Which solution will run out first? It is tempting to choose based on concentration or volume, but neither tells you how many moles are available relative to the reaction’s requirements.

The balanced equation provides the missing information.

01Why the balanced equation matters

When calcium chloride and sodium phosphate are mixed, they react to form solid calcium phosphate and dissolved sodium chloride.

The balanced chemical equation is:

\[ \ce{3CaCl2(aq) + 2Na3PO4(aq) -> Ca3(PO4)2(s) + 6NaCl(aq)} \]

Look carefully at the coefficients.

For every 3 mol of calcium chloride, the reaction requires 2 mol of sodium phosphate. These amounts produce 1 mol of calcium phosphate.

The important ratios are:

• 3 mol \(\ce{CaCl2}\) : 2 mol \(\ce{Na3PO4}\)
• 3 mol \(\ce{CaCl2}\) : 1 mol \(\ce{Ca3(PO4)2}\)
• 2 mol \(\ce{Na3PO4}\) : 1 mol \(\ce{Ca3(PO4)2}\)

The net ionic equation makes the actual chemical change clearer:

\[ \ce{3Ca^2+(aq) + 2PO4^3-(aq) -> Ca3(PO4)2(s)} \]

Calcium ions and phosphate ions combine to form an insoluble solid. Sodium and chloride ions remain dissolved.

Think of the coefficients as the ingredients required for one complete reaction set. Every set needs three calcium chloride units and two sodium phosphate units.

If one solution cannot supply enough of its required ingredient, the reaction cannot continue, even if the other solution has plenty left.

This analogy helps you understand the fixed ratio. Real reactions involve ions interacting in solution and forming an extended ionic solid, rather than ingredients being assembled into physical kits.

02What actually happens when the solutions are mixed?

Imagine setting up a school chemistry experiment.

You measure the calcium chloride solution into one graduated cylinder and the sodium phosphate solution into another. Both are transferred into the same beaker and stirred.

A cloudy white suspension forms. Some of the solid begins to settle towards the bottom.

The solid is calcium phosphate.

The experiment can then be continued by allowing the precipitate to settle, separating it from the liquid using filtration, and drying the collected solid before measuring its mass.

There are two different questions to answer:

1. Which reactant was completely consumed?
2. How much solid was produced?

The first is answered using stoichiometry. The second can be investigated by weighing the dried precipitate and comparing it with the theoretical yield.

Why cloudiness alone isn’t enough

The appearance of a white precipitate confirms that a reaction producing an insoluble solid has occurred.

However, the cloudiness does not tell you which reactant is in excess. A cloudy mixture can contain a large amount of excess calcium ions or a large amount of excess phosphate ions.

A useful experimental check is to filter the mixture and divide the clear filtrate into two separate samples.

• Add a few drops of calcium chloride solution to the first sample.
• Add a few drops of sodium phosphate solution to the second sample.

If adding calcium chloride produces more white precipitate, phosphate ions were still present in the original filtrate.

If adding sodium phosphate produces more white precipitate, calcium ions were still present.

Each added solution supplies the missing reacting ion.

For our original example, the calculation will show that calcium ions remain in excess. Therefore, adding sodium phosphate to the filtrate should produce more calcium phosphate.

This is a qualitative check. It tells you which reacting ion remains, but it does not measure the exact amount left over. The mole calculation is still needed for that.

03The method: from solution volumes to the limiting reagent

Use the following sequence whenever two solutions containing reacting substances are mixed.

Step 1: Write the balanced chemical equation

Never compare reactant amounts before balancing the equation.

The coefficients determine the mole ratio required for complete reaction.

For example:

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

This reaction requires 3 mol of A for every 2 mol of B.

Step 2: Calculate the moles in each solution

Use:

\[ n=cV \]

where:

• \(n\) is the amount of substance, in mol;
• \(c\) is the molar concentration, in mol L⁻¹;
• \(V\) is the solution volume, in L.

Convert millilitres into litres by dividing by 1000.

For example:

\[ 40.0\text{ mL}=0.0400\text{ L} \]

Calculate the initial moles of each reactant separately.

Important: Use the original volume and concentration of each solution. You do not need the total mixed volume to calculate the initial moles.

Step 3: Divide each amount by its coefficient

This is the key step.

For the general reaction:

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

calculate:

\[ \frac{n(A)}{3} \qquad \text{and} \qquad \frac{n(B)}{2} \]

The smaller value identifies the limiting reagent.

Why does this work?

Each result tells you how much reaction can occur using the available amount of that reactant.

The reactant that supports the smaller amount of reaction runs out first.

For reactions with more than two reactants, divide each reactant’s mole amount by its own balanced-equation coefficient and compare all the results.

Step 4: Calculate the theoretical product

Once you know the limiting reagent, use its mole amount and the balanced equation to calculate the maximum amount of product.

Do not calculate the theoretical yield from the excess reactant.

The reaction cannot continue once the limiting reagent has been consumed.

Step 5: Calculate the excess reactant remaining

Use:

\[ n_{\text{remaining}} = n_{\text{initial}}-n_{\text{reacted}} \]

The amount that reacts must come from the balanced equation.

Do not simply subtract the initial moles of one reactant from the initial moles of the other unless the reaction has a 1:1 mole ratio.

The calculation sequence at a glance

A useful mathematical way to organise the method is to define the maximum reaction amount as \(x\).

For a reaction:

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

the maximum reaction amount is:

\[ x=\min\left( \frac{n(A)}{a}, \frac{n(B)}{b} \right) \]

Here, \(a\), \(b\), and \(p\) are the balanced-equation coefficients, and all mole amounts are in mol.

You can then calculate:

\[ n(P)=px \]

and:

\[ n(A)_{\text{remaining}}=n(A)_{\text{initial}}-ax \]

This gives you a consistent way to calculate both product and leftover reactant.

For a broader explanation of the limiting-reagent concept, see Limiting Reagent Reactions: HSC Chemistry Explained.

06A common trap: equal moles do not mean equal reaction supply

Consider another mixture using the calcium phosphate reaction.

You mix:

• 20.0 mL of 0.150 mol L⁻¹ \(\ce{CaCl2}\);
• 20.0 mL of 0.150 mol L⁻¹ \(\ce{Na3PO4}\).

Both solutions have the same concentration and volume.

You might predict that equal moles means neither reactant is in excess.

That prediction is incorrect.

Calculate the initial amounts:

\[ n(\ce{CaCl2})=0.150\times0.0200 =0.00300\text{ mol} \]

\[ n(\ce{Na3PO4})=0.150\times0.0200 =0.00300\text{ mol} \]

Now divide by the coefficients.

For calcium chloride:

\[ \frac{0.00300}{3}=0.00100 \]

For sodium phosphate:

\[ \frac{0.00300}{2}=0.00150 \]

Calcium chloride has the smaller result.

Therefore, calcium chloride is limiting, even though both solutions initially contained exactly the same number of moles.

The amount of sodium phosphate consumed is:

\[ n(\ce{Na3PO4\ reacted}) =0.00100\times2 =0.00200\text{ mol} \]

The amount remaining is:

\[ \begin{aligned} n(\ce{Na3PO4\ remaining}) &=0.00300-0.00200\\ &=0.00100\text{ mol} \end{aligned} \]

The theoretical amount of calcium phosphate is:

\[ n(\ce{Ca3(PO4)2})=0.00100\text{ mol} \]

Its theoretical mass is:

\[ m=0.00100\times310.18 =0.310\text{ g} \]

This is why dividing by the coefficient is more reliable than simply comparing the number of moles.

For a simpler 1:1 precipitation example, see Mixing Two Solutions: A Worked Limiting-Reagent Problem.

07Common mistakes to avoid

One particularly useful check is to look at your final leftover amounts.

The limiting reagent should have zero moles remaining in an idealised complete reaction. The excess reagent should have a positive amount remaining.

If both reactants have positive amounts remaining, check your calculations and assumptions.

If both have zero remaining, you may have an exactly stoichiometric mixture.

08Questions and solutions

The following questions move from straightforward mole comparisons to more demanding combinations of solution stoichiometry, gas production, experimental evidence, and sequential additions.

Question 1

A student mixes 20.0 mL of 0.100 mol L⁻¹ copper(II) sulfate solution with 30.0 mL of 0.100 mol L⁻¹ sodium hydroxide solution.

The balanced equation is:

\[ \ce{CuSO4(aq) + 2NaOH(aq) -> Cu(OH)2(s) + Na2SO4(aq)} \]

Calculate:

a) The limiting reagent.

b) The theoretical mass of copper(II) hydroxide produced.

c) The amount of excess reactant remaining, in moles.

Use \(M(\ce{Cu(OH)2})=97.56\text{ g mol}^{-1}\).

Solution 1

Sodium hydroxide is the limiting reagent. The theoretical yield is 0.146 g of copper(II) hydroxide, and 0.000500 mol of copper(II) sulfate remains.

Step 1: Calculate the initial moles.

For copper(II) sulfate:

\[ \begin{aligned} n(\ce{CuSO4}) &=0.100\times0.0200\\ &=0.00200\text{ mol} \end{aligned} \]

For sodium hydroxide:

\[ \begin{aligned} n(\ce{NaOH}) &=0.100\times0.0300\\ &=0.00300\text{ mol} \end{aligned} \]

Step 2: Divide by the coefficients.

For copper(II) sulfate:

\[ \frac{0.00200}{1}=0.00200 \]

For sodium hydroxide:

\[ \frac{0.00300}{2}=0.00150 \]

Sodium hydroxide has the smaller value, so it is limiting.

Step 3: Calculate the theoretical product.

The equation shows a 2:1 ratio between sodium hydroxide and copper(II) hydroxide.

\[ \begin{aligned} n(\ce{Cu(OH)2}) &=0.00300\times\frac{1}{2}\\ &=0.00150\text{ mol} \end{aligned} \]

Convert to mass:

\[ \begin{aligned} m&=nM\\ &=0.00150\times97.56\\ &=0.14634\text{ g} \end{aligned} \]

Therefore:

\[ m(\ce{Cu(OH)2})=0.146\text{ g} \]

Step 4: Calculate the excess copper(II) sulfate remaining.

One mole of copper(II) sulfate reacts with two moles of sodium hydroxide.

Therefore:

\[ n(\ce{CuSO4\ reacted}) =\frac{0.00300}{2} =0.00150\text{ mol} \]

Subtract from the initial amount:

\[ \begin{aligned} n(\ce{CuSO4\ remaining}) &=0.00200-0.00150\\ &=0.000500\text{ mol} \end{aligned} \]

The final results are:

• Limiting reagent: \(\ce{NaOH}\)
• Theoretical precipitate mass: 0.146 g
• Excess reagent remaining: 0.000500 mol \(\ce{CuSO4}\)

Question 2

Sodium carbonate solution is mixed with hydrochloric acid solution.

The reaction is:

\[ \ce{Na2CO3(aq) + 2HCl(aq) -> 2NaCl(aq) + H2O(l) + CO2(g)} \]

A student mixes:

• 25.0 mL of 0.200 mol L⁻¹ \(\ce{Na2CO3}\);
• 60.0 mL of 0.150 mol L⁻¹ \(\ce{HCl}\).

Calculate:

a) The limiting reagent.

b) The amount of carbon dioxide produced, in moles.

c) The volume of carbon dioxide produced at conditions where the molar gas volume is 24.8 L mol⁻¹.

d) The amount of excess sodium carbonate remaining.

Assume complete reaction and that the measured gas is dry.

Solution 2

Hydrochloric acid is the limiting reagent. It produces 0.00450 mol of carbon dioxide, equivalent to 0.112 L at the stated conditions, while 0.000500 mol of sodium carbonate remains.

Step 1: Calculate the initial moles.

For sodium carbonate:

\[ \begin{aligned} n(\ce{Na2CO3}) &=0.200\times0.0250\\ &=0.00500\text{ mol} \end{aligned} \]

For hydrochloric acid:

\[ \begin{aligned} n(\ce{HCl}) &=0.150\times0.0600\\ &=0.00900\text{ mol} \end{aligned} \]

Step 2: Identify the limiting reagent.

Divide each amount by its coefficient.

For sodium carbonate:

\[ \frac{0.00500}{1}=0.00500 \]

For hydrochloric acid:

\[ \frac{0.00900}{2}=0.00450 \]

Hydrochloric acid has the smaller result.

Therefore, hydrochloric acid is limiting.

The maximum reaction amount is:

\[ x=0.00450\text{ mol} \]

Step 3: Calculate the amount of carbon dioxide.

The equation shows that 2 mol of hydrochloric acid produce 1 mol of carbon dioxide.

\[ \begin{aligned} n(\ce{CO2}) &=0.00900\times\frac{1}{2}\\ &=0.00450\text{ mol} \end{aligned} \]

Step 4: Calculate the gas volume.

Use:

\[ V=nV_m \]

where \(V_m\) is the molar gas volume.

\[ \begin{aligned} V(\ce{CO2}) &=0.00450\times24.8\\ &=0.1116\text{ L} \end{aligned} \]

Therefore:

\[ V(\ce{CO2})=0.112\text{ L} \]

Step 5: Calculate the excess sodium carbonate remaining.

One mole of sodium carbonate reacts with two moles of hydrochloric acid.

Therefore:

\[ \begin{aligned} n(\ce{Na2CO3\ reacted}) &=0.00900\times\frac{1}{2}\\ &=0.00450\text{ mol} \end{aligned} \]

The amount remaining is:

\[ \begin{aligned} n(\ce{Na2CO3\ remaining}) &=0.00500-0.00450\\ &=0.000500\text{ mol} \end{aligned} \]

The important point is that the carbon dioxide yield comes from the limiting reagent, not the larger initial mole amount.

Question 3

A student mixes 40.0 mL of 0.100 mol L⁻¹ sulfuric acid with 50.0 mL of 0.200 mol L⁻¹ sodium hydroxide.

Assume complete neutralisation to sodium sulfate:

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

Calculate:

a) The limiting reagent.

b) The theoretical mass of sodium sulfate formed.

c) The amount of excess sodium hydroxide remaining.

d) The final hydroxide-ion concentration, assuming additive volumes and complete dissociation of the remaining sodium hydroxide.

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

Solution 3

Sulfuric acid is the limiting reagent. The theoretical yield is 0.568 g of sodium sulfate, 0.00200 mol of sodium hydroxide remains, and the final hydroxide-ion concentration is 0.0222 mol L⁻¹.

Step 1: Calculate the initial moles.

For sulfuric acid:

\[ \begin{aligned} n(\ce{H2SO4}) &=0.100\times0.0400\\ &=0.00400\text{ mol} \end{aligned} \]

For sodium hydroxide:

\[ \begin{aligned} n(\ce{NaOH}) &=0.200\times0.0500\\ &=0.0100\text{ mol} \end{aligned} \]

Step 2: Divide by the coefficients.

For sulfuric acid:

\[ \frac{0.00400}{1}=0.00400 \]

For sodium hydroxide:

\[ \frac{0.0100}{2}=0.00500 \]

Sulfuric acid has the smaller result.

Therefore, sulfuric acid is limiting.

Step 3: Calculate the theoretical sodium sulfate yield.

The mole ratio between sulfuric acid and sodium sulfate is 1:1.

\[ n(\ce{Na2SO4})=0.00400\text{ mol} \]

Convert to mass:

\[ \begin{aligned} m&=nM\\ &=0.00400\times142.04\\ &=0.56816\text{ g} \end{aligned} \]

Therefore:

\[ m(\ce{Na2SO4})=0.568\text{ g} \]

This is the theoretical mass formed. Sodium sulfate remains dissolved in the mixture rather than forming a precipitate.

Step 4: Calculate the excess sodium hydroxide.

The equation requires 2 mol of sodium hydroxide for every 1 mol of sulfuric acid.

\[ \begin{aligned} n(\ce{NaOH\ reacted}) &=0.00400\times2\\ &=0.00800\text{ mol} \end{aligned} \]

Subtract from the initial amount:

\[ \begin{aligned} n(\ce{NaOH\ remaining}) &=0.0100-0.00800\\ &=0.00200\text{ mol} \end{aligned} \]

Step 5: Calculate the final hydroxide-ion concentration.

First calculate the total volume:

\[ V_{\text{total}}=0.0400+0.0500 =0.0900\text{ L} \]

Each mole of remaining sodium hydroxide supplies one mole of hydroxide ions.

Therefore:

\[ \begin{aligned} [\mathrm{OH}^{-}] &=\frac{0.00200}{0.0900}\\ &=0.0222\text{ mol L}^{-1} \end{aligned} \]

This final concentration is calculated using the total mixed volume, not the original sodium hydroxide volume.

Question 4

A student mixes 25.0 mL of 0.100 mol L⁻¹ magnesium chloride solution with 40.0 mL of 0.100 mol L⁻¹ sodium hydroxide solution.

The reaction is:

\[ \ce{MgCl2(aq) + 2NaOH(aq) -> Mg(OH)2(s) + 2NaCl(aq)} \]

After the reaction, the student filters the mixture and divides the clear filtrate into two separate test tubes.

The following observations are recorded:

• Adding sodium hydroxide to the first test tube produces more white precipitate.
• Adding magnesium chloride to the second test tube produces no additional precipitate.

Answer the following:

a) Identify the limiting reagent.

b) Calculate the theoretical mass of magnesium hydroxide produced.

c) Calculate the amount of excess reactant remaining.

d) Explain how the two filtrate tests support your conclusion.

Use \(M(\ce{Mg(OH)2})=58.32\text{ g mol}^{-1}\).

Solution 4

Sodium hydroxide is the limiting reagent, while magnesium chloride remains in excess. The theoretical precipitate mass is 0.117 g, and 0.000500 mol of magnesium chloride remains.

Step 1: Calculate the initial moles.

For magnesium chloride:

\[ \begin{aligned} n(\ce{MgCl2}) &=0.100\times0.0250\\ &=0.00250\text{ mol} \end{aligned} \]

For sodium hydroxide:

\[ \begin{aligned} n(\ce{NaOH}) &=0.100\times0.0400\\ &=0.00400\text{ mol} \end{aligned} \]

Step 2: Identify the limiting reagent.

Divide each amount by its coefficient.

For magnesium chloride:

\[ \frac{0.00250}{1}=0.00250 \]

For sodium hydroxide:

\[ \frac{0.00400}{2}=0.00200 \]

Sodium hydroxide has the smaller value, so it is limiting.

Step 3: Calculate the theoretical mass of magnesium hydroxide.

The equation shows a 2:1 ratio between sodium hydroxide and magnesium hydroxide.

\[ \begin{aligned} n(\ce{Mg(OH)2}) &=0.00400\times\frac{1}{2}\\ &=0.00200\text{ mol} \end{aligned} \]

Convert to mass:

\[ \begin{aligned} m&=nM\\ &=0.00200\times58.32\\ &=0.11664\text{ g} \end{aligned} \]

Therefore:

\[ m(\ce{Mg(OH)2})=0.117\text{ g} \]

Step 4: Calculate the excess magnesium chloride remaining.

One mole of magnesium chloride reacts with two moles of sodium hydroxide.

\[ \begin{aligned} n(\ce{MgCl2\ reacted}) &=\frac{0.00400}{2}\\ &=0.00200\text{ mol} \end{aligned} \]

Subtract from the initial amount:

\[ \begin{aligned} n(\ce{MgCl2\ remaining}) &=0.00250-0.00200\\ &=0.000500\text{ mol} \end{aligned} \]

Step 5: Interpret the filtrate tests.

The first test tube receives additional sodium hydroxide.

The formation of more white magnesium hydroxide means magnesium ions were still present in the filtrate. The added hydroxide ions reacted with these remaining magnesium ions.

The second test tube receives additional magnesium chloride.

No new precipitate forms, indicating that hydroxide ions were no longer present in sufficient quantity to produce more magnesium hydroxide under the idealised assumptions.

These observations support the calculated result:

• Magnesium chloride is in excess.
• Sodium hydroxide is limiting.

The reason for using separate test tubes is important. If both reagents were added to the same sample, the first addition would change the composition of the solution before the second test.

The tests identify which reacting ion remains, while the stoichiometric calculation determines the actual amount.

Question 5

A student investigates the reaction between calcium chloride and sodium phosphate:

\[ \ce{3CaCl2(aq) + 2Na3PO4(aq) -> Ca3(PO4)2(s) + 6NaCl(aq)} \]

The student initially mixes:

• 40.0 mL of 0.100 mol L⁻¹ calcium chloride;
• 30.0 mL of 0.100 mol L⁻¹ sodium phosphate.

After the initial reaction has finished, the student adds another 10.0 mL of 0.100 mol L⁻¹ calcium chloride to the same beaker.

The original precipitate is not removed before the second addition.

Assume complete reaction at each stage, negligible calcium phosphate solubility, additive volumes, and no material loss.

Determine:

a) The limiting reagent in the initial mixture.

b) The amount of sodium phosphate remaining after the initial reaction.

c) Which reactant limits the additional reaction after the second calcium chloride addition.

d) The total theoretical mass of calcium phosphate present after both additions.

e) The final amount of excess reactant remaining.

Finally, determine whether the total theoretical yield would change if all 50.0 mL of calcium chloride solution had been mixed with the 30.0 mL of sodium phosphate solution at the beginning.

Solution 5

Calcium chloride is initially limiting, but phosphate becomes limiting after the additional calcium chloride is added. The total theoretical calcium phosphate yield is 0.465 g, with 0.000500 mol of calcium chloride equivalent remaining.

The important idea is that the identity of the limiting reagent can change when more reactant is added.

Step 1: Calculate the initial moles of both reactants.

For calcium chloride:

\[ \begin{aligned} n(\ce{CaCl2}) &=0.100\times0.0400\\ &=0.00400\text{ mol} \end{aligned} \]

For sodium phosphate:

\[ \begin{aligned} n(\ce{Na3PO4}) &=0.100\times0.0300\\ &=0.00300\text{ mol} \end{aligned} \]

Step 2: Identify the initial limiting reagent.

Divide each amount by its coefficient.

For calcium chloride:

\[ \frac{0.00400}{3} =0.001333 \]

For sodium phosphate:

\[ \frac{0.00300}{2} =0.001500 \]

Calcium chloride has the smaller result.

Therefore, calcium chloride is initially limiting.

The initial reaction amount is:

\[ x_1=0.001333\text{ mol} \]

The initial amount of calcium phosphate produced is:

\[ n(\ce{Ca3(PO4)2})=0.001333\text{ mol} \]

Step 3: Calculate how much sodium phosphate remains.

The balanced equation requires 2 mol of sodium phosphate for every 3 mol of calcium chloride.

Therefore:

\[ \begin{aligned} n(\ce{Na3PO4\ reacted}) &=0.00400\times\frac{2}{3}\\ &=0.002667\text{ mol} \end{aligned} \]

Initially, there were 0.00300 mol.

Therefore:

\[ \begin{aligned} n(\ce{Na3PO4\ remaining}) &=0.00300-0.002667\\ &=0.000333\text{ mol} \end{aligned} \]

At this point, all the original calcium chloride has been consumed, but some sodium phosphate remains.

Step 4: Calculate the effect of adding more calcium chloride.

The student adds another 10.0 mL of 0.100 mol L⁻¹ calcium chloride.

The additional amount is:

\[ \begin{aligned} n(\ce{CaCl2\ added}) &=0.100\times0.0100\\ &=0.00100\text{ mol} \end{aligned} \]

We now have:

• 0.00100 mol of newly added calcium chloride;
• 0.000333 mol of sodium phosphate remaining.

Divide by the coefficients again.

For calcium chloride:

\[ \frac{0.00100}{3} =0.000333 \]

For sodium phosphate:

\[ \frac{0.000333}{2} =0.0001667 \]

Sodium phosphate now has the smaller value.

Therefore, sodium phosphate limits the additional reaction.

This is the crucial change.

Calcium chloride limited the first reaction because there was not enough calcium chloride to consume all the sodium phosphate. Adding more calcium chloride changes that balance.

Step 5: Calculate the additional precipitate formed.

The additional sodium phosphate reacts completely.

The amount of calcium phosphate produced in the second stage is:

\[ \begin{aligned} n(\ce{Ca3(PO4)2\ added}) &=0.000333\times\frac{1}{2}\\ &=0.0001667\text{ mol} \end{aligned} \]

The total amount of calcium phosphate is therefore:

\[ \begin{aligned} n(\ce{Ca3(PO4)2\ total}) &=0.001333+0.0001667\\ &=0.001500\text{ mol} \end{aligned} \]

Step 6: Calculate the total theoretical mass.

Using:

\[ M(\ce{Ca3(PO4)2})=310.18\text{ g mol}^{-1} \]

we obtain:

\[ \begin{aligned} m&=nM\\ &=0.001500\times310.18\\ &=0.46527\text{ g} \end{aligned} \]

Therefore:

\[ \boxed{m(\ce{Ca3(PO4)2})=0.465\text{ g}} \]

Step 7: Calculate the final excess calcium chloride.

The remaining sodium phosphate is completely consumed.

The amount of calcium chloride required to consume it is:

\[ \begin{aligned} n(\ce{CaCl2\ required}) &=0.000333\times\frac{3}{2}\\ &=0.000500\text{ mol} \end{aligned} \]

The second addition supplied 0.00100 mol of calcium chloride.

Therefore:

\[ \begin{aligned} n(\ce{CaCl2\ remaining}) &=0.00100-0.000500\\ &=0.000500\text{ mol} \end{aligned} \]

The final excess reactant is calcium chloride.

Step 8: Check whether the order of mixing changes the theoretical yield.

This is the most interesting part of the question.

Instead of adding the calcium chloride in two stages, imagine mixing all 50.0 mL at once with the original 30.0 mL of sodium phosphate.

The total calcium chloride amount would be:

\[ n(\ce{CaCl2})=0.100\times0.0500 =0.00500\text{ mol} \]

The sodium phosphate amount remains:

\[ n(\ce{Na3PO4})=0.00300\text{ mol} \]

Divide by the coefficients.

For calcium chloride:

\[ \frac{0.00500}{3}=0.001667 \]

For sodium phosphate:

\[ \frac{0.00300}{2}=0.001500 \]

Sodium phosphate is now limiting.

The theoretical amount of calcium phosphate is:

\[ n(\ce{Ca3(PO4)2})=0.001500\text{ mol} \]

This is exactly the same amount calculated for the two-stage addition.

The theoretical yield does not change.

The reason is that, under the stated assumptions, the total reaction depends on the total moles of each reactant and their balanced-equation ratio, not the order in which the solutions are poured together.

The tempting mistake is to treat the first limiting reagent as permanently limiting. It isn’t. The limiting reagent describes the amounts available at a particular stage of the reaction.

09What to remember when two solutions are mixed

The most reliable method is:

1. Balance the equation.
2. Calculate the initial moles of each reactant using \(n=cV\).
3. Divide each mole amount by its coefficient.
4. Identify the smallest result as the limiting reagent.
5. Use the limiting reagent to calculate the theoretical product.
6. Calculate the excess reagent by subtracting the amount consumed from its initial amount.
7. If required, convert the leftover amount into mass or final concentration.

The key skill is keeping the balanced-equation ratio involved throughout the calculation. Once you know how much reaction can occur, product yields and leftover amounts become straightforward.

The next useful step is recognising what the question wants you to do with the excess reactant: report its remaining moles, convert those moles into mass, or calculate its concentration in the final mixture. Those are different final calculations, but they all begin with the same limiting-reagent method.