As one can see from the reaction, 2 volumes of CO react with one volume of oxygen:
V(CO)/2=V(O_2)
Now, let’s calculate the partial volume of oxygen in the mixture:
V(O_2 )=200 (cm^3 )·0.21=42 (cm^3 )
Now, let’s compare the volumes of oxygen and carbon monoxide:
V(CO)/2=10(cm^3 )
V(O_2 )=42(cm^3 )
Thus, we conclude that there is an excess oxygen. Assuming that the rest of gases stay unchanged upon the reaction, we can calculate the final volume of the mixture:
V_final=V_inert+V_(O_2,excess)+V_(CO_2 ),
where V_inert is the volume of gases that are not oxygen in an initial mixture of 200cm3, V_(O_2,excess) is the excess volume of oxygen, V_(CO_2 ) is the volume of carbon dioxide formed.
V_inert=200(cm^3 )·(1-0.21)= 158(cm^3 )
V_(O_2,excess)=42(cm^3 )-10(cm^3 ) =32 (cm^3 )
V_(CO_2 )= V_CO=20 (cm^3 )
V_final= 158+32+20 (cm^3 )=210 (cm^3 )
Answer: 210 cm3
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