10 #cm^3# of carbon monoxide and 10 #cm^3# of oxygen are mixed. What will be the total volume of gas present, in #cm^3#, after the reaction?

This one can be a little tricky, and I'll explain why.

It's crucial to understand that in reactions involving gases maintained at constant pressure and temperature, the mole ratios between the species involved in the reaction equal the volume ratio.

Examine the balanced chemical equation associated with your reaction.

However, since the pressure and temperature of the gases we work with are the same, you can say the same thing about their volumes.

Stated differently, the amount of carbon monoxide that is consumed by the reaction is always twice that of the oxygen gas.

This leads you to the conclusion that you are working with a limiting reagent; in particular, carbon monoxide will do so because it will be used up before all of the oxygen has a chance to react.

Thus, it can be said that the response will only use

This implies that following the completion of the reaction, the reaction vessel will contain

According to the reaction equation (2CO + O_2 \rightarrow 2CO_2), for every 2 volumes of carbon monoxide (CO) and 1 volume of oxygen ((O_2)) consumed, 2 volumes of carbon dioxide ((CO_2)) are produced.

Given that 10 cm^3 of carbon monoxide and 10 cm^3 of oxygen are mixed, the total volume of gas present before the reaction is 20 cm^3.

After the reaction, all the carbon monoxide and oxygen will react to form carbon dioxide. Since the reaction consumes equal volumes of CO and (O_2), and the ratio of CO to (O_2) is 2:1, all of both gases will react fully. Therefore, the total volume of gas after the reaction will still be 20 cm^3.