Chemistry · General chemistry I · Worked example

Find a limiting reactant that has more moles

Methane burns in oxygen: ${CH}_{4} (g) + 2 O_{2} (g)$ $\to {CO}_{2} (g) + 2 H_{2} O (g)$ . A burner mixes $16.00 g$ of ${CH}_{4}$ with $48.00 g$ of $O_{2}$ . Which reactant limits, how much ${CO}_{2}$ can form, and how much of the excess reactant is left?

\begin{array}{l} {CH}_{4} (g) + 2 O_{2} (g) \\ \to {CO}_{2} (g) + 2 H_{2} O (g) \end{array}

Start from the balanced equation

Each ${CH}_{4}$ molecule needs two $O_{2}$ molecules. Count the atoms to be sure the ratio is right: one carbon, four hydrogen and four oxygen atoms on each side.

{CH}_{4} (g) + 2 O_{2} (g) \to {CO}_{2} (g) + 2 H_{2} O (g)

\begin{array}{c} {CH}_{4} (g) + 2 O_{2} (g) \\ \to {CO}_{2} (g) + 2 H_{2} O (g) \end{array}

Convert the methane to moles

$M ({CH}_{4})$ $= 12.01 + 4 (1.008)$ $= 16.04 g / mol$ . Dividing by it, $16.00 g$ is $0.9975 mol$ , kept to the four figures of $16.00 g$ .

16.00 g {CH}_{4} \times \frac{1 mol {CH}_{4}}{16.04 g {CH}_{4}} = 0.9975 mol {CH}_{4}

\begin{array}{c} 16.00 g {CH}_{4} \times \frac{1 mol {CH}_{4}}{16.04 g {CH}_{4}} \\ = 0.9975 mol {CH}_{4} \end{array}

Convert the oxygen to moles

$M (O_{2})$ $= 2 (16.00)$ $= 32.00 g / mol$ , so $48.00 g$ is $1.500 mol$ . There are more moles of oxygen than of methane.

48.00 g O_{2} \times \frac{1 mol O_{2}}{32.00 g O_{2}} = 1.500 mol O_{2}

\begin{array}{c} 48.00 g O_{2} \times \frac{1 mol O_{2}}{32.00 g O_{2}} \\ = 1.500 mol O_{2} \end{array}

Divide each by its coefficient

Methane’s coefficient is $1$ , so its quotient stays $0.9975$ . Oxygen’s coefficient is $2$ , so its quotient is $1.500 \div 2 = 0.7500$ . Oxygen’s quotient is smaller, so oxygen limits even though there is more of it.

\frac{1.500 mol O_{2}}{2} = 0.7500 mol

Find the theoretical yield of ${CO}_{2}$

Start from the oxygen, the limiting reactant. One ${CO}_{2}$ forms for every two $O_{2}$ , so $1.500 mol$ of $O_{2}$ gives $0.7500 mol$ of ${CO}_{2}$ , which is $33.01 g$ .

48.00 g O_{2} \times \frac{1 mol O_{2}}{32.00 g O_{2}} \times \frac{1 mol {CO}_{2}}{2 mol O_{2}} \times \frac{44.01 g {CO}_{2}}{1 mol {CO}_{2}} = 33.01 g {CO}_{2}

\begin{array}{c} 48.00 g O_{2} \times \frac{1 mol O_{2}}{32.00 g O_{2}} \\ \times \frac{1 mol {CO}_{2}}{2 mol O_{2}} \\ \times \frac{44.01 g {CO}_{2}}{1 mol {CO}_{2}} = 33.01 g {CO}_{2} \end{array}

Find the methane that reacts

The same $1.500 mol$ of $O_{2}$ uses half as many moles of methane, $0.7500 mol$ , which is $12.03 g$ .

48.00 g O_{2} \times \frac{1 mol O_{2}}{32.00 g O_{2}} \times \frac{1 mol {CH}_{4}}{2 mol O_{2}} \times \frac{16.04 g {CH}_{4}}{1 mol {CH}_{4}} = 12.03 g {CH}_{4}

\begin{array}{c} 48.00 g O_{2} \times \frac{1 mol O_{2}}{32.00 g O_{2}} \\ \times \frac{1 mol {CH}_{4}}{2 mol O_{2}} \\ \times \frac{16.04 g {CH}_{4}}{1 mol {CH}_{4}} = 12.03 g {CH}_{4} \end{array}

Subtract to find the leftover

$16.00 g - 12.03 g$ $= 3.97 g$ of methane is left unburned. The mass balances. $12.03 g$ of ${CH}_{4}$ and $48.00 g$ of $O_{2}$ react, about $60.0 g$ in all, and they form $33.01 g$ of ${CO}_{2}$ and $27.03 g$ of water.

16.00 g - 12.03 g = 3.97 g {CH}_{4}

Result

$O_{2}$ is the limiting reactant. The theoretical yield is $33.01 g$ of ${CO}_{2}$ , and $3.97 g$ of ${CH}_{4}$ is left over.

Your turn

1. One like the example

Propane burns as $C_{3} H_{8} (g) + 5 O_{2} (g)$ $\to 3 {CO}_{2} (g) + 4 H_{2} O (g)$ . A mixture holds $10.0 g$ of $C_{3} H_{8}$ and $30.0 g$ of $O_{2}$ . Which reactant limits, and how much ${CO}_{2}$ can form?

Show the answer and explanation

$O_{2}$ limits, and $24.8 g$ of ${CO}_{2}$ can form.

$10.0 g$ of propane is $0.227 mol$ and $30.0 g$ of oxygen is $0.938 mol$ . Divided by the coefficients, $0.227 \div 1 = 0.227$ and $0.938 \div 5 = 0.188$ , so oxygen limits despite its larger number of moles. Then $0.9375 mol O_{2}$ $\times 3 / 5$ $\times 44.01 g / mol$ $= 24.8 g$ of ${CO}_{2}$ .

\begin{array}{l} C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 {CO}_{2} (g) + 4 H_{2} O (g) \\ 10.0 g C_{3} H_{8} \times \frac{1 mol C_{3} H_{8}}{44.09 g C_{3} H_{8}} = 0.227 mol C_{3} H_{8} \\ 30.0 g O_{2} \times \frac{1 mol O_{2}}{32.00 g O_{2}} = 0.938 mol O_{2} \\ 30.0 g O_{2} \times \frac{1 mol O_{2}}{32.00 g O_{2}} \times \frac{3 mol {CO}_{2}}{5 mol O_{2}} \times \frac{44.01 g {CO}_{2}}{1 mol {CO}_{2}} = 24.8 g {CO}_{2} \end{array}

\begin{array}{c} C_{3} H_{8} (g) + 5 O_{2} (g) \\ \to 3 {CO}_{2} (g) + 4 H_{2} O (g) \\ 10.0 g C_{3} H_{8} \times \frac{1 mol C_{3} H_{8}}{44.09 g C_{3} H_{8}} \\ = 0.227 mol C_{3} H_{8} \\ 30.0 g O_{2} \times \frac{1 mol O_{2}}{32.00 g O_{2}} \\ = 0.938 mol O_{2} \\ 30.0 g O_{2} \times \frac{1 mol O_{2}}{32.00 g O_{2}} \\ \times \frac{3 mol {CO}_{2}}{5 mol O_{2}} \times \frac{44.01 g {CO}_{2}}{1 mol {CO}_{2}} \\ = 24.8 g {CO}_{2} \end{array}

2. A variation

Ammonia forms as $N_{2} (g) + 3 H_{2} (g) \to 2 {NH}_{3} (g)$ . Starting from $2.00 mol$ of $N_{2}$ and $5.00 mol$ of $H_{2}$ , how much ${NH}_{3}$ can form, and how much $N_{2}$ is left?

Show the answer and explanation

$3.33 mol$ of ${NH}_{3}$ forms, and $0.33 mol$ of $N_{2}$ is left.

The amounts are already in moles, so go straight to the quotients. $2.00 \div 1 = 2.00$ for $N_{2}$ and $5.00 \div 3 = 1.67$ for $H_{2}$ , so hydrogen limits. It makes $5.00 \times 2 / 3 = 3.33 mol$ of ${NH}_{3}$ and uses $5.00 \times 1 / 3 = 1.67 mol$ of $N_{2}$ , leaving $2.00 - 1.67 = 0.33 mol$ .

\begin{array}{l} N_{2} (g) + 3 H_{2} (g) \to 2 {NH}_{3} (g) \\ 5.00 mol H_{2} \times \frac{2 mol {NH}_{3}}{3 mol H_{2}} = 3.33 mol {NH}_{3} \\ 5.00 mol H_{2} \times \frac{1 mol N_{2}}{3 mol H_{2}} = 1.67 mol N_{2} \\ 2.00 mol - 1.67 mol = 0.33 mol \end{array}

\begin{array}{c} N_{2} (g) + 3 H_{2} (g) \to 2 {NH}_{3} (g) \\ 5.00 mol H_{2} \times \frac{2 mol {NH}_{3}}{3 mol H_{2}} \\ = 3.33 mol {NH}_{3} \\ 5.00 mol H_{2} \times \frac{1 mol N_{2}}{3 mol H_{2}} \\ = 1.67 mol N_{2} \\ 2.00 mol - 1.67 mol = 0.33 mol \end{array}

3. Somewhere new

$50.0 mL$ of $0.100 M {AgNO}_{3}$ is mixed with $30.0 mL$ of $0.200 M NaCl$ , and $AgCl$ precipitates: ${AgNO}_{3} (aq) + NaCl (aq)$ $\to AgCl (s) + {NaNO}_{3} (aq)$ . What mass of $AgCl$ can form?

Show the answer and explanation

$0.717 g$ of $AgCl$ .

In solution, moles are molarity times volume in liters. ${AgNO}_{3}$ gives $0.0500 L \times 0.100 mol / L$ $= 0.00500 mol$ and $NaCl$ gives $0.00600 mol$ . They react one to one, so ${AgNO}_{3}$ limits. $0.00500 mol \times 143.32 g / mol$ $= 0.717 g$ of $AgCl$ .

\begin{array}{l} {AgNO}_{3} (aq) + NaCl (aq) \to AgCl (s) + {NaNO}_{3} (aq) \\ n = (0.0500 L) (0.100 mol/L) = 0.00500 mol \\ n = (0.0300 L) (0.200 mol/L) = 0.00600 mol \\ 0.00500 mol {AgNO}_{3} \times \frac{1 mol AgCl}{1 mol {AgNO}_{3}} \times \frac{143.32 g AgCl}{1 mol AgCl} = 0.717 g AgCl \end{array}

\begin{array}{c} {AgNO}_{3} (aq) + NaCl (aq) \\ \to AgCl (s) + {NaNO}_{3} (aq) \\ n = (0.0500 L) (0.100 mol/L) \\ = 0.00500 mol \\ n = (0.0300 L) (0.200 mol/L) \\ = 0.00600 mol \\ 0.00500 mol {AgNO}_{3} \\ \times \frac{1 mol AgCl}{1 mol {AgNO}_{3}} \times \frac{143.32 g AgCl}{1 mol AgCl} \\ = 0.717 g AgCl \end{array}

Keep exploring

Open the mixture in Stoichiometry & yield and raise the oxygen to $70.00 g$ : methane becomes the limiting reactant, and the ${CO}_{2}$ rises to $43.90 g$ .

Return to the concept →

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Start from the balanced equation

Convert the methane to moles

Convert the oxygen to moles

Divide each by its coefficient

Find the theoretical yield of CO2CO2​

Find the methane that reacts

Subtract to find the leftover

Result

Your turn

1. One like the example

O2O2​ limits, and 24.8 g24.8 g of CO2CO2​ can form.

2. A variation

3.33 mol3.33 mol of NH3NH3​ forms, and 0.33 mol0.33 mol of N2N2​ is left.

3. Somewhere new

0.717 g0.717 g of AgClAgCl.

Keep exploring

Study it, then try it

Find the theoretical yield of ${CO}_{2}$

$O_{2}$ limits, and $24.8 g$ of ${CO}_{2}$ can form.

$3.33 mol$ of ${NH}_{3}$ forms, and $0.33 mol$ of $N_{2}$ is left.

$0.717 g$ of $AgCl$ .