Help needed please? Thanks!?

Question

A mixture of 10 cm^3 of methane and 10 cm^3 of ethane was mixed with an excess of oxygen and the mixture was ignited. After cooling to room temperature, the residual gas was passed through a solution of potassium hydroxide. What was the volume of the gas absorbed by the alkali?

(1) 15 cm^3
(2) 20 cm^3
(3) 25 cm^3
(4) 30 cm^3

The answer is (4) 30 cm^3...I don't understand...Can you show me the workings and with explanation too? Thanks a lot.

Answer 1

The combustion of the hydrocarbons yields CO2 and water.

CH4 + 2O2 == CO2 + 2H2O

2C2H6 + 7O2 == 4CO2 + 6H2O

So you will get 3 molar quantities of CO2 for the combined 2 molar quantities of methane and ethane. So 10 cm3 each will yield 30 cm3 of CO2 gas to be absorbed in the KOH

Answer 2

CH4 + 2O2 ----> CO2 +2H2O

So by volume 10 cm^3 of methane produces 10 cm^3 of CO2 You have a 1 : 1 ratio by volume

2C2H6 + 7 O2 -----> 4CO2 + 6H2O

So, again by volume, 10 cm^3 of ethane produces twice that volume of CO2 or 20 cm^. You have a 2 : 1 ratio by volume.

So the total volume of CO2 produced is 10 cm^3 from the methane and 20 cm^3 from the ethane, a total of 30 cm^3

Answer 3

CH4 + 2O2 --> CO2 + 2H2O
2C2H6 + 7O2 --> 4CO2 + 6H2O
10 cm^3 CH4 --> 10 cm^3 CO2 (1 molecule CH4 --> 1 molecule CO2)
10 cm^3 C2H6 --> 20 cm^3 CO2 (1 molecule C2H6 --> 2 molecules CO2)