# Question #12a91

May 16, 2017

$2.07 g {C}_{3} {H}_{8}$

#### Explanation:

First, let's find the moles of ${O}_{2}$ using the ideal-gas equation:

$m o l {O}_{2} = \frac{\left(1.04 a t m\right) \left(5.53 L\right)}{\left(0.08206 \frac{L - a t m}{m o l - K}\right) \left(298 K\right)} = 0.235 m o l {O}_{2}$

Now, we'll use the stoichiometrically equivalent values from the equation to find the relative number of moles of ${C}_{3} {H}_{8}$:

$0.235 m o l {O}_{2} \left(\frac{1 m o l {C}_{3} {H}_{8}}{5 m o l {O}_{2}}\right) = 0.0470 m o l {C}_{3} {H}_{8}$

Finally, we'll use the molar mass of propane to find the mass in grams of propane needed:

$0.0470 m o l {C}_{3} {H}_{8} \left(\frac{44.11 g {C}_{3} {H}_{8}}{1 m o l {C}_{3} {H}_{8}}\right) = 2.07 g {C}_{3} {H}_{8}$