# Question #ad087

Dec 15, 2016

#### Answer:

13.6 L of $C {O}_{2}$

#### Explanation:

First, you need the balanced equation:

${C}_{3} {H}_{8}$ + 5 ${O}_{2}$ ----> 3 $C {O}_{2}$ + 4 ${H}_{2} O$

Convert 8.92 g into moles of propane (molar mass = 44 g)

8.92 / 44 = 0.2027 mol

According to the equation, 3 mol of $C {O}_{2}$ is produced for each 1 mol of propane that reacts. So, in this problem, we obtain 0.608 mol of $C {O}_{2}$.

To find the volume at STP, multiply this by molar volume (22.4 L)

0.608 x 22.4 = 13.6 L