Chemistry · General chemistry I · Worked example

Find the mass of a reactant needed

Ammonia forms from nitrogen and hydrogen, $N_{2} (g) + 3 H_{2} (g) \to 2 {NH}_{3} (g)$ . What mass of hydrogen is needed to make $50.0 g$ of ammonia?

50.0 g {NH}_{3} \to ? g H_{2}

Write the balanced equation

The first example ran forward, from a reactant to a product. This one runs backward, from the product wanted to a reactant needed. Check the balance first. Each side has two nitrogen and six hydrogen atoms.

N_{2} (g) + 3 H_{2} (g) \to 2 {NH}_{3} (g)

Find the molar masses

Ammonia is $14.01 + 3 (1.008)$ $= 17.03 g / mol$ . Hydrogen gas is diatomic, so its molar mass is $2 (1.008) = 2.016 g / mol$ , not $1.008 g / mol$ .

M ({NH}_{3}) = 14.01 + 3 (1.008) = 17.03 g/mol

\begin{array}{c} M ({NH}_{3}) = 14.01 + 3 (1.008) \\ = 17.03 g/mol \end{array}

Convert the ammonia to moles

Divide $50.0 g$ by $17.03 g / mol$ , written as a factor so grams of ${NH}_{3}$ cancel. That is $2.94 mol$ of ammonia.

50.0 g {NH}_{3} \times \frac{1 mol {NH}_{3}}{17.03 g {NH}_{3}} = 2.94 mol {NH}_{3}

\begin{array}{c} 50.0 g {NH}_{3} \times \frac{1 mol {NH}_{3}}{17.03 g {NH}_{3}} \\ = 2.94 mol {NH}_{3} \end{array}

Apply the mole ratio, wanted over given

Hydrogen is wanted and ammonia is given, so the ratio is $3 mol H_{2}$ over $2 mol {NH}_{3}$ . Moles of ${NH}_{3}$ cancel. Every $2 mol$ of ammonia needs $3 mol$ of hydrogen, so the amount grows by half, to $4.40 mol$ .

2.94 mol {NH}_{3} \times \frac{3 mol H_{2}}{2 mol {NH}_{3}} = 4.40 mol H_{2}

\begin{array}{c} 2.94 mol {NH}_{3} \times \frac{3 mol H_{2}}{2 mol {NH}_{3}} \\ = 4.40 mol H_{2} \end{array}

Convert to grams of hydrogen

Multiply by $2.016 g / mol$ . Carry the whole chain from the $50.0 g$ so that rounding happens once. $4.404 \times 2.016 = 8.88 g$ .

4.40 mol H_{2} \times \frac{2.016 g H_{2}}{1 mol H_{2}} = 8.88 g H_{2}

\begin{array}{c} 4.40 mol H_{2} \times \frac{2.016 g H_{2}}{1 mol H_{2}} \\ = 8.88 g H_{2} \end{array}

Check by conserving mass

Hydrogen supplies only part of the ammonia’s mass, and nitrogen supplies the rest. $2.936 mol$ of ${NH}_{3}$ needs $1.468 mol$ of $N_{2}$ , which is $41.1 g$ . $8.88 g + 41.1 g = 50.0 g$ , the mass of ammonia made.

Result

$8.88 g$ of hydrogen is needed to make $50.0 g$ of ammonia.

Your turn

1. One like the example

What mass of nitrogen is needed to make the same $50.0 g$ of ammonia?

Show the answer and explanation

$41.1 g$ of $N_{2}$ .

The chain is the same with nitrogen as the wanted substance. $50.0 g \div 17.03 g / mol$ $= 2.936 mol$ of ${NH}_{3}$ . The ratio is $1 mol N_{2}$ per $2 mol {NH}_{3}$ , which gives $1.468 mol$ , and $1.468 \times 28.02 = 41.1 g$ .

\begin{array}{l} N_{2} (g) + 3 H_{2} (g) \to 2 {NH}_{3} (g) \\ 50.0 g {NH}_{3} \times \frac{1 mol {NH}_{3}}{17.03 g {NH}_{3}} \times \frac{1 mol N_{2}}{2 mol {NH}_{3}} \times \frac{28.02 g N_{2}}{1 mol N_{2}} = 41.1 g N_{2} \end{array}

\begin{array}{c} N_{2} (g) + 3 H_{2} (g) \to 2 {NH}_{3} (g) \\ 50.0 g {NH}_{3} \times \frac{1 mol {NH}_{3}}{17.03 g {NH}_{3}} \\ \times \frac{1 mol N_{2}}{2 mol {NH}_{3}} \\ \times \frac{28.02 g N_{2}}{1 mol N_{2}} = 41.1 g N_{2} \end{array}

2. A variation

Methane burns by ${CH}_{4} (g) + 2 O_{2} (g)$ $\to {CO}_{2} (g) + 2 H_{2} O (l)$ . What mass of water forms from $8.00 g$ of methane?

Show the answer and explanation

$18.0 g$ of $H_{2} O$ .

This runs forward, from a reactant to a product. $8.00 g \div 16.04 g / mol$ $= 0.4988 mol$ of ${CH}_{4}$ . Each ${CH}_{4}$ gives two $H_{2} O$ , so $0.9975 mol$ of water forms, and $0.9975 \times 18.02 = 18.0 g$ .

\begin{array}{l} {CH}_{4} (g) + 2 O_{2} (g) \to {CO}_{2} (g) + 2 H_{2} O (l) \\ 8.00 g {CH}_{4} \times \frac{1 mol {CH}_{4}}{16.04 g {CH}_{4}} \times \frac{2 mol H_{2} O}{1 mol {CH}_{4}} \times \frac{18.02 g H_{2} O}{1 mol H_{2} O} = 18.0 g H_{2} O \end{array}

\begin{array}{c} {CH}_{4} (g) + 2 O_{2} (g) \\ \to {CO}_{2} (g) + 2 H_{2} O (l) \\ 8.00 g {CH}_{4} \times \frac{1 mol {CH}_{4}}{16.04 g {CH}_{4}} \\ \times \frac{2 mol H_{2} O}{1 mol {CH}_{4}} \\ \times \frac{18.02 g H_{2} O}{1 mol H_{2} O} = 18.0 g H_{2} O \end{array}

3. Somewhere new

Zinc dissolves in acid by $Zn (s) + 2 HCl (aq)$ $\to {ZnCl}_{2} (aq) + H_{2} (g)$ . What volume of $0.500 M HCl$ dissolves $2.50 g$ of zinc?

Show the answer and explanation

$0.153 L$ , or $153 mL$ , of the acid.

The chain ends in moles of $HCl$ , $2.50 g \div 65.38 g / mol \times 2$ $= 0.0765 mol$ . A molarity then turns moles into a volume, $V = n / M$ $= 0.0765 mol \div 0.500 M$ $= 0.153 L$ .

\begin{array}{l} Zn (s) + 2 HCl (aq) \to {ZnCl}_{2} (aq) + H_{2} (g) \\ 2.50 g Zn \times \frac{1 mol Zn}{65.38 g Zn} \times \frac{2 mol HCl}{1 mol Zn} = 0.0765 mol HCl \\ V = \frac{n}{M} \\ n = 0.0765 mol \\ M = 0.500 M \\ V = \frac{0.0765 mol}{0.500 M} = 0.153 L \end{array}

\begin{array}{c} Zn (s) + 2 HCl (aq) \\ \to {ZnCl}_{2} (aq) + H_{2} (g) \\ 2.50 g Zn \times \frac{1 mol Zn}{65.38 g Zn} \\ \times \frac{2 mol HCl}{1 mol Zn} = 0.0765 mol HCl \\ V = \frac{n}{M} \\ n = 0.0765 mol \\ M = 0.500 M \\ V = \frac{0.0765 mol}{0.500 M} = 0.153 L \end{array}

Keep exploring

Open the chain in a Chemistry box, change $50.0 g$ to $100. g$ of ${NH}_{3}$ , and predict the hydrogen before the checker confirms $17.8 g$ .

Return to the concept →

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After reading

Study it, then try it

Practice the key terms as flashcards, then open the example in its tool, change a value and keep a useful result on your board.

Study the 4 key terms Check the chain in a Chemistry box Open worked example on a board Periodic table and atomic masses

Your existing work stays on this device. Examples open as editable copies.

Write the balanced equation

Find the molar masses

Convert the ammonia to moles

Apply the mole ratio, wanted over given

Convert to grams of hydrogen

Check by conserving mass

Result

Your turn

1. One like the example

41.1 g41.1 g of N2N2​.

2. A variation

18.0 g18.0 g of H2OH2​O.

3. Somewhere new

0.153 L0.153 L, or 153 mL153 mL, of the acid.

Keep exploring

Study it, then try it

$41.1 g$ of $N_{2}$ .

$18.0 g$ of $H_{2} O$ .

$0.153 L$ , or $153 mL$ , of the acid.