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F2.4 · Solve stoichiometry problems involving gases

Learn to solve stoichiometry problems involving gases through clear examples and targeted practice.

Ontario Grade 11 Chemistry

Gases and Atmospheric Chemistry

Use balanced equations to connect gas volume, amount, and reacting quantities

A balloon gets larger when more gas enters it. The added gas consists of particles that spread out and occupy space. Stoichiometry connects the amount of those particles to the amounts of substances in a balanced reaction. In this lesson, you will use that connection to find an unknown gas amount or volume. The gas conditions matter: a volume must be linked to its temperature and pressure.

What you will learn

1. From the balanced equation to a mole ratio

A balanced chemical equation shows the relative amounts of reactants used and products formed. The numbers in front of the formulas are called coefficients. They give mole ratios, not mass ratios or volume ratios in every situation.
For example, the balanced reaction between hydrogen gas and oxygen gas forms water. The coefficients show that two moles of hydrogen react with one mole of oxygen to form two moles of water. The atom count is the same on both sides: four hydrogen atoms and two oxygen atoms.
A mole is a counting unit for particles. In a stoichiometry problem, first use the balanced equation to find the mole ratio between the known substance and the unknown substance. If a gas volume is given, convert it to an amount in moles before using that ratio, unless the gases are at the same temperature and pressure.
2H2(g)+O2(g)→2H2O(l)2\mathrm{H_2(g)}+\mathrm{O_2(g)}\rightarrow2\mathrm{H_2O(l)}

2. Converting gas volume to amount

Gas particles are far apart compared with particles in liquids and solids. As a result, a gas sample can change volume when its temperature or pressure changes. Temperature describes how hot or cold the gas is. Pressure describes the force of gas particles against the walls of their container.
At a stated reference condition, a course table may give the molar volume: the volume occupied by one mole of gas. For example, at STP, the course convention commonly uses 22.4 litres per mole. Use the value and conditions specified in the question or course reference table. Do not apply a reference molar volume to a gas at different conditions without a stated reason.
For gas conditions that are not represented by a supplied molar volume, the ideal gas relationship connects pressure, volume, amount, and absolute temperature. Use pressure in kilopascals, volume in litres, and temperature in kelvins with the matching gas constant. Convert degrees Celsius to kelvins by adding 273.15. This is a course-level calculation tool for the gas amount; it does not change the mole ratios in the balanced equation.
When two gases are at the same temperature and pressure, equal volumes contain equal amounts in moles. In that specific case, their volume ratio matches the coefficients in the balanced equation. If conditions differ, first determine the amount of each gas using the appropriate gas relationship.
n=VVmPV=nRTn=\frac{V}{V_m}\qquad PV=nRT

3. A reliable method for gas stoichiometry

A gas stoichiometry problem follows the same core path as other stoichiometry problems: convert the given quantity to moles, use the balanced equation, then convert the result to the requested quantity. Write the units beside each value. Units help show whether each step is moving toward the requested answer.
First, identify what is given and what is asked for. Note the gas conditions and the units. Next, write and balance the reaction. Convert the given gas volume to moles using molar volume when the stated conditions match, or use the ideal gas relationship when the problem provides pressure, volume, and temperature.
Use the coefficient ratio to find the amount of the target substance. If the target is a gas volume, convert its moles to volume using the appropriate gas relationship and conditions. If a mass is involved, use the molar mass to convert between grams and moles. Molar mass is the mass of one mole of a substance, found from its chemical formula and a periodic table.
Round only the final result. Keep extra digits during intermediate calculations, then report an answer that reflects the least precise measured value. This is significant-figure practice. Include the final unit and, when useful, the gas conditions.
given quantity→n→mole ratio→requested quantity\text{given quantity}\rightarrow n\rightarrow\text{mole ratio}\rightarrow\text{requested quantity}

4. Check the result

Before accepting an answer, check that the reaction is balanced and that the coefficient ratio is written in the correct direction. A ratio for reactant to product must match the two substances named in the question.
Check the conditions used for the gas conversion. A volume at STP cannot be treated as though it were at another temperature or pressure unless the problem gives a conversion method. Also check that the final unit answers the question: an amount in moles is not a volume in litres.
A useful reasonableness check is to compare the result with the coefficients and the supplied quantity. For example, if a reaction produces two moles of gas for every mole of a reactant, the product amount should reflect that ratio. This comparison checks the setup; it does not replace the calculation.

Worked example

Finding a gas volume from a reactant amount

At STP, what volume of hydrogen gas is needed to react completely with 3.0 mol of oxygen gas? Assume the course molar volume at STP is 22.4 L/mol.
  1. Balance and identify the ratio
    The balanced reaction shows that two moles of hydrogen react with one mole of oxygen. The question gives oxygen and asks for hydrogen, so use the hydrogen-to-oxygen coefficient ratio.
    2H2(g)+O2(g)→2H2O(l)2\mathrm{H_2(g)}+\mathrm{O_2(g)}\rightarrow2\mathrm{H_2O(l)}
  2. Find the hydrogen amount
    Multiply the known oxygen amount by the coefficient ratio. The oxygen units cancel, leaving moles of hydrogen.
    3.0 mol O2×2 mol H21 mol O2=6.0 mol H23.0\ \mathrm{mol\ O_2}\times\frac{2\ \mathrm{mol\ H_2}}{1\ \mathrm{mol\ O_2}}=6.0\ \mathrm{mol\ H_2}
  3. Convert amount to volume
    The requested gas is at STP, so use the stated molar volume. Moles cancel, leaving litres.
    6.0 mol H2×22.4 L1 mol=134.4 L H26.0\ \mathrm{mol\ H_2}\times\frac{22.4\ \mathrm{L}}{1\ \mathrm{mol}}=134.4\ \mathrm{L\ H_2}
Answer: The required volume is 1.3×102 L H21.3\times10^2\ \mathrm{L\ H_2} at STP. The result is rounded to two significant figures because 3.0 mol has two significant figures.
Check: The balanced equation requires twice as many moles of hydrogen as oxygen. The calculation gives 6.0 mol of hydrogen for 3.0 mol of oxygen, so the ratio is correct. Multiplying by 22.4 L/mol gives a volume in litres.

Common mistakes and how to avoid them

Using the coefficients as a direct ratio between grams and litres.
Correction: Coefficients give mole ratios. Convert the known quantity to moles first, then use the ratio.
Using the STP molar volume for any gas volume in any conditions.
Correction: Use a molar volume only when the problem's conditions match the stated reference conditions.
Reversing the coefficient ratio.
Correction: Write the coefficient for the wanted substance on top and the coefficient for the known substance on the bottom.
Rounding each intermediate value.
Correction: Keep extra digits through the calculation and round the final answer to appropriate significant figures.

Lesson summary

Check your understanding

Question 1

At the same temperature and pressure, a reaction uses 1 mol of oxygen for every 2 mol of hydrogen. What is the hydrogen-to-oxygen gas volume ratio?
  1. 1:2
  2. 2:1
  3. 1:1
  4. 4:1
Show answer and explanation
2:1
At the same temperature and pressure, gas volume ratios match mole ratios. The equation's hydrogen-to-oxygen ratio is 2:1.

Question 2

A gas sample has volume 11.2 L at STP. Using 22.4 L/mol, how many moles of gas are present?
  1. 0.25 mol
  2. 0.50 mol
  3. 2.0 mol
  4. 11.2 mol
Show answer and explanation
0.50 mol
Divide the volume by the molar volume: 11.2 L divided by 22.4 L/mol equals 0.50 mol.

Question 3

A problem gives a gas volume at conditions that do not match a listed molar volume, along with its pressure and temperature. Which relationship can find its amount?
  1. n=VVmn=\frac{V}{V_m} using the STP molar volume without checking conditions
  2. PV=nRTPV=nRT with consistent units and temperature in kelvins
  3. The balanced equation alone, without converting to moles
  4. The mass-to-mole conversion, even when no mass is given
Show answer and explanation
PV=nRTPV=nRT with consistent units and temperature in kelvins
The ideal gas relationship uses the stated pressure, volume, and absolute temperature to find the amount. The STP molar volume should not be used when the conditions do not match.

Key terms

Mole ratio
A comparison of amounts of substances taken from the coefficients in a balanced chemical equation.
Molar volume
The volume occupied by one mole of a gas at stated conditions.
Ideal gas relationship
A relationship connecting gas pressure, volume, amount, and absolute temperature.
Significant figures
Digits that show the precision of a measured or calculated value.

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About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Chemistry (SCH3U), expectation F2.4. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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