Gas-Volume Calculations
Use a stated molar gas volume and conditions, convert cm³ and dm³, and connect gas amount to a balanced-equation ratio.
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The core idea
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Learning objectives
- calculate stoichiometric reacting masses and volumes of gases (one mole of gas occupies 24 dm3 at room temperature and pressure); calculations involving the idea of limiting reactants may be set (knowledge of the gas laws and the calculations of gaseous volumes at different temperatures and pressures are not required)
1. Outcome and prerequisites
By the end, you should be able to use a stated molar gas volume at stated conditions, convert cm³ and dm³ correctly, and connect gas volume, amount and a balanced-equation ratio.
Keep V and Vₘ in matching volume units. 1 dm³ = 1000 cm³.
Temperature and pressure affect the volume occupied per mole, so use only the stated molar gas volume. The balanced-equation coefficients describe the chemical ratio and do not change with gas conditions.
2. Modelled example
Modelled example 1
Convert a solid mass to a gas volume
Problem
View solution step by step
Establish the relationship
Method
Write and balance the thermal decomposition equation.Reason
The equation establishes how carbonate amount relates to carbon-dioxide amount.Working
CaCO₃(s) → CaO(s) + CO₂(g)Convert the supplied solid mass
Method
Divide the carbonate mass by its molar mass.Reason
The equation ratio uses amount, so mass must become moles.Working
n(CaCO₃) = 5.00/100.0 = 0.0500 mol.Apply the ratio
Method
Transfer carbonate amount to carbon-dioxide amount using coefficients.Reason
The coefficient ratio CaCO₃:CO₂ is 1:1.Working
n(CO₂) = 0.0500 mol.Use the stated gas condition
Method
Multiply gas amount by the stated molar gas volume, then convert the requested unit.Reason
The stated conditions determine volume per mole; multiplying by 1000 converts cubic decimetres to cubic centimetres.Working
V = 0.0500(24.0) = 1.20 dm³ = 1.20 × 10³ cm³.
3. Guided practice
Guided practice 2
Keep gas units consistent across a ratio
Problem
State why a direct ratio is valid
Hints
Hint 1: shared conditions
Hint 2: coefficient direction
View solution step by step
Use the same-condition volume ratio
Method
Multiply hydrogen volume by ammonia coefficient over hydrogen coefficient.Reason
At the same conditions, gas volumes are proportional to amounts, so the 3:2 coefficient ratio can act directly on the two gas volumes.Working
V(NH₃) = 600(2/3) = 400 cm³.
4. Plausible error contrast
A learner uses 600 cm³/24.0 dm³ mol⁻¹ = 25.0 mol. The first invalid step is mixing volume units. Either convert 600 cm³ = 0.600 dm³ and calculate 0.600/24.0 = 0.0250 mol, or use 24000 cm³ mol⁻¹ directly.
5. Changed-context transfer
Mind stretcher 1: Use a molar volume that is not 24 dm³ mol⁻¹Extension
At a stated temperature and pressure, the molar gas volume is 25.0 dm³ mol⁻¹. For 2KClO₃(s) → 2KCl(s) + 3O₂(g), find the oxygen volume from 0.0800 mol potassium chlorate.
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The 2:3 ratio gives n(O₂) = 0.0800(3/2) = 0.120 mol. Use the stated value, not a memorised RTP value: V = 0.120(25.0) = 3.00 dm³.
6. Independent evidence
Magnesium reacts with hydrochloric acid to form magnesium chloride and hydrogen. Calculate the hydrogen volume at stated conditions where Vₘ = 24.0 dm³ mol⁻¹ when 12.15 g magnesium reacts with excess acid. Use Mg = 24.3. Show the balanced equation with states, conversion to amount, coefficient ratio, gas-volume conversion, final unit and a magnitude check. Then express the answer in cm³.
Continue to Solution Concentration.