# How do you calculate the volume of 66.5 g of carbon monoxide at STP?

Jul 3, 2018

$\text{Volume} = 53.2 {\mathrm{dm}}^{3}$

#### Explanation:

Well simple comparison method..

$C O \rightarrow S T P$

$\text{Molar mass of} \textcolor{w h i t e}{x} C O = 12 + 16 = 28 g m o {l}^{-} 1$

Let the volume be represented as $x$

Therefore;

$28 g m o {l}^{-} 1 \rightarrow 22.4 m o l {\mathrm{dm}}^{-} 3$

$66.5 g \rightarrow x {\mathrm{dm}}^{3}$

Cross multiplying;

$28 \times x = 22.4 \times 66.5$

$28 x = 1489.6$

$x = \frac{1489.6}{28}$

$x = 53.2 {\mathrm{dm}}^{3}$