40 cm3 of methane (CH4) is reacted with 60 cm3 of oxygen. The equation for the reaction is given below:
CH4 + 2O2 ⟶ CO2 + 2H2O
All volumes are measured at room temperature. What is the total volume of the gases remaining at the end of the reaction?
- (a)60 cm3
- (b)40 cm3
- (c)45 cm3
- (d)50 cm3
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40 cm3
Reason — From the equation :
CH4 + 2O2 ⟶ CO2 + 2H2O
1 volume of CH4 reacts with 2 volumes of O2
For 40 cm3 of CH4 the required O2 is 40 cm3 x 2 = 80 cm3
Since only 60 cm3 of O2 is available.
The amount of CH4 that reacts with 60 cm3 of O2 is = $\frac{60}{2}$ = 30 cm3
∴ The excess CH4 remaining is 40 cm3 - 30 cm3 = 10 cm3
From the balanced equation, 2 volumes of O2 produce 1 volume of CO2.
∴ The volume of CO2 produced from 60 cm of O2 is $\frac{60}{2}$ = 30 cm3
The remaining gases are excess CH4 and produced CO2.
∴ Total volume = 10 cm3 (excess CH4) + 30 cm3 (CO2) = 40 cm3