# What is the volume of 36.8 g of carbon monoxide at STP?

May 9, 2016

$V = 29.4 L$

#### Explanation:

Let us first find the number of mole of carbon monoxide in $36.8 g$:

$n = \frac{m}{M M} = \frac{36.8 \cancel{g}}{28.0 \frac{\cancel{g}}{m o l}} = 1.31 m o l$

At STP, the molar volume is equal to $22.4 \frac{L}{m o l}$, therefore, the volume that will be occupied by $36.8 g$ carbon monoxide is:

$V = 1.31 \cancel{m o l} \times \frac{22.4 L}{1 \cancel{m o l}} = 29.4 L$