#2CO(g) + O_2(g) -> 2CO_2(g)#. How many liters of #O_2# will react with 74.2 liters of CO? If 6.5 g of CO react, how many liters of #CO_2# are produced?

1 Answer
Apr 9, 2017

Answer:

37.1 L of #O_2#.
5.20 L of #CO_2#

Explanation:

We calculate the relative quantities from the molar ratios in the balanced reaction equation.

#2CO(g)+O_2(g)→2CO_2(g)#.
As a pretty good approximation, gas volumes are proportional to molar amounts. Therefore with a molar ratio of 2:1 of #CO : O_2# we can calculate that 74.2 L of CO will react with
#(74.2/2) = 37.1# L of #O_2#.

Similarly, converting the CO mass to moles, we have #(6.5g)/(28"g/mol") = 0.232# moles CO.

This is approximately 22.4L /mol * 0.232 moles = 5.20 L at “STP” - slightly more a “NTP”.
This would produce the same molar and volumetric amount of #CO_2#.