Question

`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?

Answer 1

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`.