Hard water contains dissolved calcium and magnesium ions, $\mathrm{Ca^{2+}}$ and $\mathrm{Mg^{2+}}$, and leaves a white crust in kettles. Washing soda, sodium carbonate, $\mathrm{Na_2\!CO_3}$, softens such water by precipitating these ions as solid carbonates, so they are removed from the water. Milk of magnesia is a white solid made by mixing two clear solutions. In each case two solutions of ionic compounds are mixed and an insoluble solid, a precipitate, forms. How much solid forms is a stoichiometry question, with molarity and volume taking the place of a weighed mass.

In this lesson you will go from the volume and molarity of a solution to the mass of precipitate, and find which solution runs out first when two solutions are mixed (Figure 1).

1. Volume of solutionmL, then L
2. $\times C$molarity, mol per L
3. Moles of reactant
4. $\times \dfrac{c_\mathrm{P}}{c_\mathrm{A}}$mole ratio from the equation
5. Moles of precipitatetwo solutions: the smaller amount
6. $\times M$molar mass of the precipitate
7. Mass of precipitateg

$n = C \times V$ with V in L.$m = n \times M$.

Learning Objectives

After this lesson, you should be able to:

1. Calculate the mass of precipitate formed from solution volumes and concentrations.
2. Find the limiting reactant when two solutions are mixed.

Key Ideas

1. What Precipitates

When two solutions of soluble ionic compounds are mixed, a precipitate forms if a cation from one and an anion from the other make an insoluble compound. The solubility guidelines (Figure 2) tell you which pair that is. Mixing calcium chloride and sodium carbonate solutions gives insoluble calcium carbonate; the sodium and chloride ions stay dissolved (Figure 3):

The net ionic equation keeps only the ions that form the solid:

Predicting the products of a reaction is covered in Classifying Chemical Reactions. Here the balanced equation is given, and the question is how much solid forms.

Key idea: a precipitate forms when a cation and an anion from the two solutions make an insoluble compound. In this lesson the precipitates are calcium carbonate, $\mathrm{CaCO_3}$, and magnesium hydroxide, $\mathrm{Mg(OH)_2}$.

2. Amount of Reactant From Molarity and Volume

In Molarity you met $C = \frac{n}{V}$, the moles of solute in each L of solution. Rearranged, it gives the moles of a reactant in a measured volume:

Volumes are usually measured in mL, so convert first: $1\,\text{L} = 1000\,\text{mL}$. A molarity of 0.185 M means 0.185 mol of solute in each liter, so $0.185\,\text{mol/L}$ is a unit factor, just like a molar mass.

Key idea: moles of a dissolved reactant $= C \times V$, with $V$ in L. This replaces the grams-to-moles step of a mass-to-mass problem.

3. The Mole Ratio, Then the Mass of Precipitate

From moles of reactant, the steps are the ones you used in Mass-to-Mass Stoichiometry: the mole ratio from the balanced equation (wanted over given), then the molar mass of the precipitate. For magnesium hydroxide made from sodium hydroxide, 2 mol of NaOH give 1 mol of $\mathrm{Mg(OH)_2}$:

The molar masses used in this lesson are $\mathrm{CaCO_3}$: $100.09\,\text{g/mol}$ and $\mathrm{Mg(OH)_2}$: $58.32\,\text{g/mol}$.

Key idea: use the coefficients of the balanced equation to turn moles of reactant into moles of precipitate, then multiply by the molar mass of the precipitate: $m = n \times M$.

4. Two Solutions Mixed: the Limiting Reactant

When volumes of both solutions are given, one of them usually runs out first. As in Limiting Reactant and Theoretical Yield, find the moles of precipitate each reactant could form. The smaller amount is what actually forms, and that reactant is the limiting reactant; the other is in excess.

Do not compare the moles of the two reactants directly: in Worked Example 3 the potassium hydroxide solution holds more moles than the magnesium sulfate solution, yet it runs out first, because 2 mol of KOH are needed for each mole of $\mathrm{Mg(OH)_2}$ (Figure 4).

Key idea: with two solutions, compute the moles of precipitate from each reactant. The smaller amount wins; its reactant is limiting.

Try it: mix two solutions and find the limiting reactant

Pick a reaction, set the volume and molarity of each solution, and see which reactant runs out first and how much precipitate forms.

5. A Strategy for Precipitation Problems

Follow the roadmap in Figure 1:

1. Write the balanced equation and identify the precipitate.
2. Convert each volume to liters and find moles: $n = C \times V$.
3. Use the mole ratio to get moles of precipitate. With two solutions, do this for each reactant and keep the smaller amount.
4. Convert to grams with the molar mass of the precipitate. Keep 4 significant figures until the last step and round the answer to 3.

Exam tip: the dilution equation $C_1V_1 = C_2V_2$ describes one solution before and after adding water. It has no coefficients in it, so it cannot describe two different substances reacting. Always go through moles and the balanced equation.

Lab note: in the lab the solid is filtered, washed, dried and weighed. The mass you calculate is the most that can form (the theoretical yield); a measured mass is usually a little less.

Worked Example 1: Calcium Carbonate From Hard Water

Problem: To model softening hard water with washing soda, excess sodium carbonate solution is added to 25.0 mL of 0.185 M calcium chloride. What mass of calcium carbonate precipitates?

Step 1. Convert the volume to liters.

Step 2. Find the moles of calcium chloride: $n = C \times V$.

Step 3. Use the mole ratio. 1 mol of $\mathrm{CaCl_2}$ gives 1 mol of $\mathrm{CaCO_3}$. Sodium carbonate is in excess, so calcium chloride decides the amount.

Step 4. Convert moles of calcium carbonate to grams.

Final Answer: 0.463 g of CaCO₃

Less than half a gram of solid from 25 mL: precipitates usually come out in small, easily weighed amounts.

1. Volume in L$V = 0.0250\,\text{L}$Step 1
2. $\times 0.185\,\text{mol/L}$Step 2: molarity of CaCl₂
3. Moles of CaCl₂$n = 0.004625\,\text{mol}$
4. $\times \dfrac{1}{1}$Step 3: 1 CaCO₃ per 1 CaCl₂
5. Moles of CaCO₃$n = 0.004625\,\text{mol}$
6. $\times 100.09\,\text{g/mol}$Step 4: molar mass of CaCO₃
7. Mass of CaCO₃$m = 0.4629\,\text{g} \approx 0.463\,\text{g}$

Your turn: mass of precipitate

Same four steps as Worked Example 1: liters, moles from $n = C \times V$, the mole ratio, then grams.

Worked Example 2: Magnesium Hydroxide From Sodium Hydroxide

Problem: To make a magnesium hydroxide antacid suspension, 30.0 mL of 0.245 M sodium hydroxide is added to excess magnesium chloride solution. What mass of magnesium hydroxide forms?

Step 1. Convert the volume to liters.

Step 2. Find the moles of sodium hydroxide.

Step 3. Use the mole ratio. 2 mol of NaOH give 1 mol of $\mathrm{Mg(OH)_2}$, so the moles are halved.

Step 4. Convert to grams.

Final Answer: 0.214 g of Mg(OH)₂

The coefficient 2 in front of NaOH is the whole difference from Worked Example 1: skip it and the answer doubles.

1. Volume in L$V = 0.0300\,\text{L}$Step 1
2. $\times 0.245\,\text{mol/L}$Step 2: molarity of NaOH
3. Moles of NaOH$n = 0.007350\,\text{mol}$
4. $\times \dfrac{1}{2}$Step 3: 1 Mg(OH)₂ per 2 NaOH
5. Moles of Mg(OH)₂$n = 0.003675\,\text{mol}$half the moles of NaOH
6. $\times 58.32\,\text{g/mol}$Step 4: molar mass of Mg(OH)₂
7. Mass of Mg(OH)₂$m = 0.2143\,\text{g} \approx 0.214\,\text{g}$

Your turn: a mole ratio of 2 : 1

Same four steps as Worked Example 2: watch the coefficient of the hydroxide in the mole ratio.

Worked Example 3: Two Solutions Mixed

Problem: A volume of 45.0 mL of 0.132 M magnesium sulfate is mixed with 40.0 mL of 0.235 M potassium hydroxide. Which reactant is limiting, and what mass of magnesium hydroxide can form?

Step 1. Convert both volumes to liters.

Step 2. Find the moles of each reactant.

Step 3. Find the moles of magnesium hydroxide each reactant could form.

Step 4. The smaller amount wins. Potassium hydroxide gives the smaller amount (0.004700 mol), so KOH is the limiting reactant and magnesium sulfate is in excess.

Step 5. Convert the smaller amount to grams.

Final Answer: KOH is limiting; 0.274 g of Mg(OH)₂

Some magnesium sulfate is left over in the solution. How much is the subject of the lesson on the excess reactant remaining.

1. Moles of each reactant$n_{\mathrm{MgSO_4}} = 0.005940\,\text{mol}$ $n_{\mathrm{KOH}} = 0.009400\,\text{mol}$Steps 1 and 2
2. Step 3: mole ratio for each
3. Moles of Mg(OH)₂ possible$\text{from}\ \mathrm{MgSO_4}:\ 0.005940\,\text{mol}$ $\text{from}\ \mathrm{KOH}:\ 0.004700\,\text{mol}$KOH gives less: limiting
4. $\times 58.32\,\text{g/mol}$Steps 4 and 5: keep the smaller amount
5. Mass of Mg(OH)₂$m = 0.2741\,\text{g} \approx 0.274\,\text{g}$

Your turn: two solutions mixed

Same five steps as Worked Example 3: moles of each reactant, moles of precipitate from each, keep the smaller, then grams.

Common Mistakes

Throughout, keep 4 significant figures in the moles and the unrounded mass, and round only the final answer to 3.

Common mistake: Leaving the volume in mL. In Worked Example 1, multiplying 0.185 mol/L by 25.0 (in mL) gives 462.9 g of calcium carbonate from 25 mL of solution, 1000 times the right answer. Molarity is in mol/L, so convert mL to L first.

Common mistake: Leaving out the coefficients. In Worked Example 2, taking the moles of $\mathrm{Mg(OH)_2}$ equal to the moles of NaOH gives 0.429 g, twice the right answer of 0.214 g.

Common mistake: Turning the mole ratio upside down. In Worked Example 2, multiplying by 2 instead of dividing gives 0.857 g, four times the right answer. Put the substance you want on top: $\frac{1\,\text{mol}\;\mathrm{Mg(OH)_2}}{2\,\text{mol}\;\mathrm{NaOH}}$.

Common mistake: Comparing the moles of the reactants. In Worked Example 3, magnesium sulfate has fewer moles (0.005940 against 0.009400 mol), so it looks limiting. Using it gives 0.346 g, about 26% too much. Compare the moles of precipitate each reactant can form.

Common mistake: Using the dilution equation for a reaction. $C_1V_1 = C_2V_2$ is for diluting one solution. For two solutions that react, go through moles and the mole ratio of the balanced equation.

Key Takeaway

For a precipitation reaction in solution, find the moles of each dissolved reactant from $n = C \times V$ (with $V$ in L), use the mole ratio of the balanced equation to get moles of the precipitate, and multiply by its molar mass. When two solutions are mixed, compute the moles of precipitate from each reactant: the smaller amount forms, and its reactant is the limiting reactant.

Practice: randomized problems

Each problem below uses fresh numbers every time and mixes every kind of problem in this lesson: work it out, check your answer, then open the worked solution to compare your steps.

Your results

This summary collects every practice problem you try on this page and suggests what to review next.

Figures: original diagrams by Dr. Ray’s Learning Center, licensed CC BY 4.0. Data: Whitten, Davis, Peck and Stanley, Chemistry, 9th ed. (Brooks/Cole, Cengage Learning, 2010), Section 3-8 (pp. 107 to 109) for moles of solute from volume and molarity and the reaction ratio; Section 6-9 (pp. 238 to 239) for precipitation reactions; the solubility guidelines and Table 6-4 (p. 217) for which compounds are soluble. Molar masses: the practice engine’s molar-mass tables (IUPAC atomic weights). Simulator: original work by Dr. Ray’s Learning Center, licensed CC BY 4.0; reactions, solubility and colors from Whitten 9th ed. (pp. 16, 107 to 109, 217, 238 to 239, 849), molar masses from the practice engine’s IUPAC tables. The practice problems are generated and checked by an audited engine; see the lesson audit log.