# Question e05a9

Jan 24, 2016

At STP, 13.0 L of methane has a mass of 9.18 g.

#### Explanation:

You will need the molar volume of a gas at STP

At STP of $\text{273.15 K}$ and $\text{100 kPa}$, the molar volume of a gas is $\text{22.711 L/mol}$.

You will need to determine the moles of methane by dividing the volume in liters of methane by its molar volume at STP.

$13.0 \cancel{\text{L CH"_4xx (1"mol CH"_4)/(22.711 cancel"L CH"_4)="0.57241 mol CH"_4}}$ (I am keeping some guard digits to reduce rounding errors.)

In order to determine the mass of 13 L of methane at STP, you will need to multiply the moles of methane times its molar mass.

Molar mass of $\text{CH"_4}$ is $\text{16.04246 g/mol}$
http://pubchem.ncbi.nlm.nih.gov/compound/297section=Top

$0.57241 \cancel{\text{mol CH"_4xx(16.04246"g CH"_4)/(1cancel"mol CH"_4)="9.18 g CH"_4}}$ (rounded to three significant figures)