# 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?

Apr 9, 2017

37.1 L of ${O}_{2}$.
5.20 L of $C {O}_{2}$

#### 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 $C O : {O}_{2}$ we can calculate that 74.2 L of CO will react with
$\left(\frac{74.2}{2}\right) = 37.1$ L of ${O}_{2}$.

Similarly, converting the CO mass to moles, we have $\frac{6.5 g}{28 \text{g/mol}} = 0.232$ 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 $C {O}_{2}$.