# Question #71444

Apr 14, 2017

33.62 liters

#### Explanation:

Use stockhiyometry:

1 mol methane is 16 grams, but you have 12 grams. Therefore you have 0.75 mol of methane to burn.

If you burn 0.75 moles of methane, you will get 1.5 moles of water vapor. Or in other words, you will get 27 grams of water vapor.

Use ideal gas law (STP =1 atm pressure and 0 degrees C (temperature)):

$P \cdot V = n \cdot R \cdot T$
$1 \cdot V = 1.5 \cdot 0.082056 \cdot 273.15$
$V = 33.62$ liters