# What volume does 22.4 g of chlorine gas occupy at STP?

Jun 15, 2016

$7.08 L$

#### Explanation:

Because we are at STP, we must use the ideal gas law equation
$P \times V = n \times R \times T$.

• P represents pressure (must have units of atm)
• V represents volume (must have units of liters)
• n represents the number of moles
• R is the proportionality constant (has units of $\frac{L \times a t m}{m o l \times K}$)
• T represents the temperature, which must be in Kelvins.

Next, list your known and unknown variables. Our only unknown is the volume of $C {l}_{2} \left(g\right)$. Our known variables are P,n,R, and T.

At STP, the temperature is 273K and the pressure is 1 atm. The proportionality constant, R, is equal to 0.0821 $\frac{L \times a t m}{m o l \times K}$
The only issue is the mass of $C {l}_{2} \left(g\right)$, we need to convert it into moles of $C {l}_{2} \left(g\right)$ in order to use the formula.
22.4cancel"g" xx (1molCl_2)/(70.90cancel"g") = 0.316 mol $C {l}_{2}$

Now all we have to do is rearrange the equation and solve for V like so:
$V = \frac{n \times R \times T}{P}$
$V = \frac{0.316 m o l \times 0.0821 \frac{L \times a t m}{m o l \times K} \times \left(273 K\right)}{1 a t m}$
$V = 7.08 L$