Gas Stoichiometry

General Chemistry · Gas Laws and Ideal Gas Calculations

Dr. Ray’s Learning Center · covers: Gas Stoichiometry

25–35 min
7 figures
1 simulator
4 worked examples
1 practice type: gas stoichiometry

Before this lesson:

Related (optional):

Related: Partial Pressures and Dalton’s Law

An air bag has to fill in a few hundredths of a second, and the engineer who designs it must know how many grams of the solid that makes the gas to pack inside. A brewer, a baker and a chemist heating limestone ask the same question in different forms: how much gas does a reaction give, or how much reactant does a given volume of gas need?

This lesson joins two tools you already have. Stoichiometry turns grams of one substance into moles of another with a mole ratio from the balanced equation. The ideal gas law turns moles of a gas into a volume, or a volume into moles. Put together, they connect grams to liters of gas.

Grams of Athe given substance
divide by Mmolar mass of A
Moles of Ain mol
mole ratiofrom the balanced equation
Moles of gas Bin mol
ideal gas lawor 22.414 L/mol at STP
Liters of gas Bat the stated T and P

The mole ratio applies to amounts (mol) only.The chain also runs backwards, from liters to grams.

Figure 1. From grams to liters of gas: grams to moles, the mole ratio, then the ideal gas law.

Learning Objectives

After this lesson, you should be able to:

Combine reaction stoichiometry with PV = nRT to find gas volumes produced or required.

Key Ideas

1. The Gas Stoichiometry Chain

Every problem in this lesson follows the same chain. Change the grams of the given substance to moles with its molar mass. Change those moles to moles of the gas with the mole ratio from the balanced equation. Then change the moles of gas to a volume (Figure 2).

Grams to moles: divide by the molar mass.
Moles to moles: multiply by the mole ratio, wanted over given.
Moles of gas to liters: use the ideal gas law, or the molar volume at STP.

Grams of Ain g
$\times \frac{1}{M}$ molar mass of A
Moles of Ain mol
$\times \frac{\text{mol wanted}}{\text{mol given}}$ coefficients of the equation
Moles of gas Bin mol
$\times \frac{RT}{P}$ T in K, R to match P
Volume of Bin L

Figure 2. Three factors take grams of one substance to liters of a gas: one over the molar mass, the mole ratio, then RT over P.

Key idea: the mole ratio connects the amounts of two different substances, so apply it to amounts (mol), never to masses or volumes.

2. STP Shortcut or PV = nRT

At any temperature and pressure, solve the ideal gas law for the volume: $V = \dfrac{nRT}{P}$ . At standard temperature and pressure the shortcut is the molar volume, $V_{\text{m}}$ , the volume of 1 mol of an ideal gas. This lesson follows Whitten STP, 273.15 K and 1 atm. From the exact value of R:

V_{\text{m}} = \dfrac{RT}{P} \approx 22.414\,\text{L/mol}

So at STP, multiply the moles by 22.414 L/mol, or use $V = \dfrac{nRT}{P}$ with 0.08206 and 273.15 K: both routes give the same 3 s.f. answer for every problem in this lesson. You may see 22.711 L/mol in other books: that is the IUPAC standard, 273.15 K and 1 bar (100 kPa). Review: The Ideal Gas Law

Figure 3. At STP the molar volume is a shortcut for RT over P. Away from STP, only the ideal gas law gives the right volume.

Key idea: use 22.414 L/mol only at STP. At any other temperature or pressure, use $V = \dfrac{nRT}{P}$ .

Try it: from grams to liters of gas

Pick a reaction, set the mass of the reactant and the conditions, and follow the chain from grams to moles to liters. Load the settings of a worked example, switch between STP and other conditions, or run the chain backwards from a volume to a mass.

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3. From a Gas Volume Back to Grams

The chain also runs backwards. When the volume of gas is given, find its moles with the ideal gas law, use the mole ratio to get the moles of the other substance, then multiply by its molar mass (Figure 4):

n = \dfrac{PV}{RT}

Volume of gas Bin L, with T and P
$\times \frac{P V}{R T}$ n from PV over RT
Moles of gas Bin mol
$\times \frac{\text{mol wanted}}{\text{mol given}}$ coefficients of the equation
Moles of Ain mol
$\times M$ molar mass of A
Grams of Ain g

Figure 4. Backwards, the same three steps in reverse order: moles of gas from the ideal gas law, the mole ratio, then times the molar mass.

Key idea: backwards, the steps are the same in reverse: volume to moles of gas, mole ratio, then moles to grams.

Worked Example 1: Gas Volume at STP

Problem: Heating limestone decomposes calcium carbonate. What volume of CO₂, measured at STP (273.15 K, 1 atm), is produced from 25.0 g of CaCO₃?

\mathrm{CaCO_3}(\text{s}) \rightarrow \mathrm{CaO}(\text{s}) + \mathrm{CO_2}(\text{g})

Step 1. Convert the mass of CaCO₃ to moles.

M_{\mathrm{CaCO_3}} = 100.09\,\text{g/mol}

n_{\mathrm{CaCO_3}} = \dfrac{25.0\,\text{g}}{100.09\,\text{g/mol}} \approx 0.2498\,\text{mol}

Step 2. Use the mole ratio. One mole of CaCO₃ gives one mole of CO₂:

n_{\mathrm{CO_2}} = 0.2498\,\text{mol} \times \dfrac{1\,\text{mol}\ \mathrm{CO_2}}{1\,\text{mol}\ \mathrm{CaCO_3}} = 0.2498\,\text{mol}

Step 3. At STP, multiply by the molar volume.

\begin{aligned} V &= 0.2498\,\text{mol} \times 22.414\,\dfrac{\text{L}}{\text{mol}} \\ &\approx 5.599\,\text{L} \approx 5.60\,\text{L} \end{aligned}

Final Answer: 5.60 L of CO₂

Rounded operand: Step 3 uses 0.2498 mol, rounded to 4 significant figures. The unrounded amount, and the route through $V = \dfrac{nRT}{P}$ , give the same final answer, 5.60 L.

Your turn: gas volume at STP

Same three steps as Worked Example 1: grams to moles, mole ratio, then times 22.414 L/mol.

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Worked Example 2: Gas Volume at Other Conditions

Problem: Magnesium reacts with hydrochloric acid. What volume of H₂ forms from 1.50 g of Mg at 23.0 °C and 0.985 atm?

\mathrm{Mg}(\text{s}) + 2\,\mathrm{HCl}(\text{aq}) \rightarrow \mathrm{MgCl_2}(\text{aq}) + \mathrm{H_2}(\text{g})

Step 1. Convert the mass of Mg to moles, then use the mole ratio. One mole of Mg gives one mole of H₂:

n_{\mathrm{Mg}} = \dfrac{1.50\,\text{g}}{24.31\,\text{g/mol}} \approx 0.06170\,\text{mol}

n_{\mathrm{H_2}} = 0.06170\,\text{mol} \times \dfrac{1\,\text{mol}\ \mathrm{H_2}}{1\,\text{mol}\ \mathrm{Mg}} = 0.06170\,\text{mol}

Step 2. Convert the temperature to kelvin.

T = (23.0 + 273.15)\,\text{K} = 296.15\,\text{K}

Step 3. These are not STP conditions, so use $V = \dfrac{nRT}{P}$ . With the pressure in atm, use R in L·atm:

\begin{aligned} V &= \dfrac{(0.06170\,\text{mol})(0.08206\,\dfrac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}})(296.15\,\text{K})}{0.985\,\text{atm}} \\ &\approx 1.522\,\text{L} \approx 1.52\,\text{L} \end{aligned}

Final Answer: 1.52 L of H₂

Magnesium $1.50\,\text{g}\ \mathrm{Mg}$ given
$\times \frac{1\,\text{mol}}{24.31\,\text{g}}$ g to mol
Moles of Mg $0.06170\,\text{mol}\ \mathrm{Mg}$ 4 significant figures
$\times \frac{1\,\text{mol}\ \mathrm{H_2}}{1\,\text{mol}\ \mathrm{Mg}}$ mole ratio 1 to 1
Moles of H₂ $0.06170\,\text{mol}\ \mathrm{H_2}$ 4 significant figures
$\times \frac{(0.08206)(296.15\,\text{K})}{0.985\,\text{atm}}$ not STP: RT over P
Volume of H₂ $1.52\,\text{L}$ Final answer to 3 significant figures

Figure 5. Worked Example 2 one step at a time. The gas is not at STP, so the last step uses RT over P, not 22.414 L/mol.

Rounded operand: Step 3 uses 0.06170 mol, rounded to 4 significant figures. The unrounded amount gives the same final answer, 1.52 L.

Your turn: gas volume at other conditions

Same steps as Worked Example 2: grams to moles, mole ratio, kelvin, then $V = \dfrac{nRT}{P}$ .

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Worked Example 3: Volume of a Gas Reactant

Problem: Propane burns in oxygen. What volume of O₂, at 25.0 °C and 748 torr, is needed to burn 8.80 g of C₃H₈ completely?

\mathrm{C_3H_8}(\text{g}) + 5\,\mathrm{O_2}(\text{g}) \rightarrow 3\,\mathrm{CO_2}(\text{g}) + 4\,\mathrm{H_2O}(\text{l})

Step 1. Convert the mass of propane to moles.

n_{\mathrm{C_3H_8}} = \dfrac{8.80\,\text{g}}{44.10\,\text{g/mol}} \approx 0.1995\,\text{mol}

Step 2. Use the mole ratio. Each mole of propane needs five moles of O₂:

n_{\mathrm{O_2}} = 0.1995\,\text{mol} \times \dfrac{5\,\text{mol}\ \mathrm{O_2}}{1\,\text{mol}\ \mathrm{C_3H_8}} = 0.9975\,\text{mol}

Step 3. Convert the temperature to kelvin and use $V = \dfrac{nRT}{P}$ . With the pressure in torr, use R in L·torr:

T = (25.0 + 273.15)\,\text{K} = 298.15\,\text{K}

\begin{aligned} V &= \dfrac{(0.9975\,\text{mol})(62.36\,\dfrac{\text{L}\cdot\text{torr}}{\text{mol}\cdot\text{K}})(298.15\,\text{K})}{748\,\text{torr}} \\ &\approx 24.79\,\text{L} \approx 24.8\,\text{L} \end{aligned}

Final Answer: 24.8 L of O₂

Figure 6. In Worked Example 3 the oxygen needed is five times the moles of propane. Using a 1 to 1 ratio would give only one fifth of the volume.

Rounded operand: Step 2 uses 0.1995 mol, rounded to 4 significant figures. The unrounded amount gives the same final answer, 24.8 L.

Your turn: volume of a gas reactant

Same steps as Worked Example 3: the mole ratio comes from the coefficients of the balanced equation.

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Worked Example 4: Mass of Reactant for a Volume of Gas

Problem: An air bag inflates with N₂ from sodium azide. What mass of NaN₃ is needed to fill a 45.0 L bag with N₂ at 115 kPa and 30.0 °C?

2\,\mathrm{NaN_3}(\text{s}) \rightarrow 2\,\mathrm{Na}(\text{s}) + 3\,\mathrm{N_2}(\text{g})

Step 1. Convert the temperature to kelvin and find the moles of N₂. With the pressure in kPa, use R in L·kPa:

T = (30.0 + 273.15)\,\text{K} = 303.15\,\text{K}

\begin{aligned} n_{\mathrm{N_2}} &= \dfrac{(115\,\text{kPa})(45.0\,\text{L})}{(8.314\,\dfrac{\text{L}\cdot\text{kPa}}{\text{mol}\cdot\text{K}})(303.15\,\text{K})} \\ &\approx 2.053\,\text{mol} \end{aligned}

Step 2. Use the mole ratio. Three moles of N₂ come from two moles of NaN₃:

n_{\mathrm{NaN_3}} = 2.053\,\text{mol} \times \dfrac{2\,\text{mol}\ \mathrm{NaN_3}}{3\,\text{mol}\ \mathrm{N_2}} \approx 1.369\,\text{mol}

Step 3. Convert the moles of NaN₃ to grams.

\begin{aligned} m &= 1.369\,\text{mol} \times 65.01\,\text{g/mol} \\ &\approx 89.00\,\text{g} \approx 89.0\,\text{g} \end{aligned}

Final Answer: 89.0 g of NaN₃

Volume of N₂ $45.0\,\text{L}\ \mathrm{N_2}$ given
$\times \frac{115\,\text{kPa}}{(8.314)(303.15\,\text{K})}$ n from PV over RT
Moles of N₂ $2.053\,\text{mol}\ \mathrm{N_2}$ 4 significant figures
$\times \frac{2\,\text{mol}\ \mathrm{NaN_3}}{3\,\text{mol}\ \mathrm{N_2}}$ mole ratio 2 to 3
Moles of NaN₃ $1.369\,\text{mol}\ \mathrm{NaN_3}$ 4 significant figures
$\times \frac{65.01\,\text{g}}{1\,\text{mol}}$ mol to g
Mass of NaN₃ $89.0\,\text{g}$ Final answer to 3 significant figures

Figure 7. Worked Example 4 runs the chain backwards: moles of nitrogen from the ideal gas law, the mole ratio with sodium azide on top, then grams.

Rounded operands: Steps 2 and 3 carry 2.053 mol and 1.369 mol, each rounded to 4 significant figures. The unrounded amounts give the same final answer, 89.0 g.

Your turn: mass of reactant for a volume of gas

Same steps as Worked Example 4: $n = \dfrac{PV}{RT}$ for the gas, mole ratio, then times the molar mass.

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Common Mistakes

Common mistake: Using 22.414 L/mol away from STP. In Worked Example 2, 0.06170 mol times 22.414 L/mol gives 1.38 L. The gas is at 23.0 °C and 0.985 atm, not at STP, so use $V = \dfrac{nRT}{P}$ : 1.52 L.

Common mistake: Using the temperature in °C. Putting 23.0 instead of 296.15 into Worked Example 2 gives 0.118 L. Gas laws need kelvin.

Common mistake: Leaving out the mole ratio. In Worked Example 3, using the moles of propane for the O₂ gives 4.96 L, five times too little. Every mole of propane needs five moles of O₂.

Common mistake: Turning the mole ratio upside down. In Worked Example 3, dividing by 5 instead of multiplying gives 0.992 L. In Worked Example 4, using 3 over 2 instead of 2 over 3 gives about $2.00\times10^{2}\,\text{g}$ of NaN₃. Put the substance you want on top.

Common mistake: Stopping at moles. In Worked Example 4, 1.369 mol of NaN₃ is not the answer: the question asks for a mass. Multiply by 65.01 g/mol.

Exam tip: write the units on every factor. The mole ratio must cancel the moles of the given substance and leave the moles of the wanted one; R must match the pressure unit.

Key Takeaway

Gas stoichiometry is ordinary stoichiometry with the ideal gas law at one end. Grams change to moles with the molar mass, moles of one substance to moles of another with the mole ratio, and moles of gas to a volume with $V = \dfrac{nRT}{P}$ , or with 22.414 L/mol at STP only. Run the chain backwards to get a mass from a volume of gas.

The relations to keep at hand:

PV = nRT

V_{\text{m}} = \dfrac{RT}{P} \approx 22.414\,\text{L/mol}

Practice: randomized problems

Each problem below uses fresh numbers every time: work it out, check your answer, then open the worked solution to compare your steps.

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Your results

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

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Figures and simulator: original work by Dr. Ray’s Learning Center, licensed CC BY 4.0. Atomic weights: IUPAC Commission on Isotopic Abundances and Atomic Weights, standard atomic weights 2021 (abridged values), the same table as the other stoichiometry lessons. Gas constant: CODATA 2022 recommended values (exact R in SI units, rounded to 0.08206, 62.36 and 8.314 in the lesson); the molar volume 22.414 L/mol at 273.15 K and 1 atm follows from it. Gas stoichiometry: K. Whitten, R. Davis, M. Peck and G. Stanley, General Chemistry, 9th ed., section 12-12, and J. Moore, C. Stanitski and P. Jurs, Chemistry: The Molecular Science, 5th ed., section 8-5. The simulator and the practice problems are audited; see the lesson audit log.