# How do you calculate the volume of 25.0 g of carbon monoxide at STP?

Jul 16, 2016

The volume is 20.3 L.

#### Explanation:

We can convert the mass of carbon monoxide to moles and then use the Ideal Gas Law to calculate the volume at STP.

Moles of $\text{CO}$

$\text{Moles of CO" = 25.0 color(red)(cancel(color(black)("g CO"))) × ("1 mol CO")/(28.01 color(red)(cancel(color(black)("g CO")))) = "0.8925 mol CO}$

Volume at STP

The Ideal Gas Law is:

color(blue)(|bar(ul(PV = nRT)|),

where

• $P$ is the pressure
• $V$ is the volume
• $n$ is the number of moles
• $R$ is the gas constant
• $T$ is the temperature

We can rearrange the Ideal Gas Law to get

$V = \frac{n R T}{P}$

STP is 1 bar and 0 °C.

$n = \text{0.8925 mol}$
$R = \text{0.083 14 bar·L·K"^"-1""mol"^"-1}$
$T = \text{273.15 K}$
$P = \text{1 bar}$

V = (nRT)/P = (0.8925 color(red)(cancel(color(black)("mol"))) × "0.083 14" color(red)(cancel(color(black)("bar")))·"L"· color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 273.15 color(red)(cancel(color(black)("K"))))/(1 color(red)(cancel(color(black)("bar")))) = "20.3 L"