Consider the combustion of methanol at some high temperature in a constant-pressure reaction chamber:
2CH3OH (g) + 3O2 (g) ⟶2CO2 (g) + 4H2O (g)
2 L of CH3OH will react with 3 L of O2 to produce 2 L CO2 and 4 L H2O – all in gaseous state.
18L CH3OH will require 18*3/2 = 27L of O2 gas
The CH3OH is in excess – the 10L O2 will be completely consumed.
10L O2 will react with 10*2/3 = 6.67L CH3OH to produce 6.67L of CO2 and 10*4/3 = 13.33L H2O
There will be 18-6.67 = 11.33 L of CH3OH unreacted
The vessel will contain: 6.67L CO2 , 13.33L H2O and 11.33L CH3OH = 31.33L gases in total
From your choices – answer: 31.3L
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