To find which gas is left over, compare the supplied volumes using the coefficients of the balanced equation. If 40 cm³ of carbon monoxide reacts with 30 cm³ of oxygen, the 2 to 1 ratio means 40 cm³ of carbon monoxide uses 20 cm³ of oxygen. Carbon monoxide limits the reaction and 10 cm³ of oxygen remains.
The 2027 SEC G3 Chemistry K324 syllabus includes reacting gas volumes and limiting reactants. The worked example is a paper calculation. All gas volumes are compared at the same temperature and pressure, with the reaction assumed to go to completion for the limiting reactant.
- Balance the equation before comparing volumes.
- Find how much of the second gas the first gas needs.
- Use the limiting reactant to calculate product volume.
- Subtract the volume used from the starting volume for the excess gas.
CHAPTER 1 OF 4 · Back to contents
Use the balanced equation to find the volume ratio
The equation is 2CO(g) + O₂(g) → 2CO₂(g). The coefficients give a mole ratio of 2 to 1 to 2. For gases at the same temperature and pressure, this is also a volume ratio. The units can be cm³ throughout because the calculation compares like volumes.
The gas with the smaller starting volume can be in excess. Here the reaction needs only one volume of oxygen for every two volumes of carbon monoxide. Use that requirement to decide whether the oxygen supply is sufficient.
CHAPTER 2 OF 4 · Back to contents
Worked example | Find the gas mixture remaining
The question gives 40 cm³ of carbon monoxide and 30 cm³ of oxygen. To use all the carbon monoxide requires 40 × 1/2 = 20 cm³ of oxygen. There is enough oxygen, so the carbon monoxide is the limiting reactant. The oxygen left over is 30 − 20 = 10 cm³.
The equation gives two volumes of carbon dioxide for every two volumes of carbon monoxide consumed. Hence 40 cm³ of carbon dioxide is formed. The final mixture contains 40 cm³ of carbon dioxide and 10 cm³ of oxygen, for a total of 50 cm³ under the same conditions.
| Gas | Initial volume | Change | Final volume |
|---|---|---|---|
| Carbon monoxide | 40 cm³ | −40 cm³ | 0 cm³ |
| Oxygen | 30 cm³ | −20 cm³ | 10 cm³ |
| Carbon dioxide | 0 cm³ | +40 cm³ | 40 cm³ |
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CHAPTER 3 OF 4 · Back to contents
Practise | Give the reaction less oxygen
What happens with 40 cm³ of carbon monoxide and 15 cm³ of oxygen?
Using all 40 cm³ of carbon monoxide would require 20 cm³ of oxygen, but only 15 cm³ is present. Oxygen is limiting. Its 15 cm³ reacts with 30 cm³ of carbon monoxide and makes 30 cm³ of carbon dioxide. The mixture contains 10 cm³ of unreacted carbon monoxide and 30 cm³ of carbon dioxide.
What if the oxygen volume is exactly 20 cm³?
The gases are supplied in the exact 2 to 1 ratio, so both are fully consumed in this model. The final mixture contains 40 cm³ of carbon dioxide only.
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CHAPTER 4 OF 4 · Back to contents
Check what the question means by final volume
Why can the total gas volume decrease?
The balanced equation conserves atoms. It uses three moles of gaseous reactants to form two moles of gaseous product. At the same temperature and pressure, fewer moles of gas occupy a smaller volume.
Do I need 24 dm³ per mole?
For this same-condition volume ratio, the conversion factor cancels. Use the molar gas volume when converting between moles and a gas volume at the stated conditions. If the question says a gas was removed or a product condensed, apply that information before summing the gases that remain.
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Sources and next reading
SEAB 2027 SEC G3 Chemistry K324 syllabus
G3 Chemistry worked guide to theoretical and percentage yield