# How many grams of water are produced when propane (C3H8) burns with 12.0 L of oxygen at STP?

Dec 15, 2015

$7.72 \text{g H"_2"O}$

#### Explanation:

The NIE for the complete combustion of propane is:

$\text{C"_3 "H"_8 + 5 "O"_2 -> 3 "CO"_2 + 4 "H"_2"O}$

The problem doesn't state how much propane is burned, so we're going to assume that there is enough to use up all the oxygen. To find out how much that is, we need to convert liters of ${\text{O}}_{2}$ to moles of ${\text{O}}_{2}$ using the fact that at STP, a mole of gas is equivalent to 22.4 liters.

12.0 cancel("L") * (1 "mol O"_2)/(22.4 cancel("L")) = 0.536 "mol O"_2

The molar ratio of ${\text{O}}_{2}$ to $\text{H"_2"O}$ is $\frac{5}{4}$ and the molecular weight of $\text{H"_2"O}$ is $18.0 \text{g}$, so:

0.536 cancel("mol O"_2) * (4 cancel("mol H"_2"O"))/(5 cancel("mol O"_2)) * (18.0"g H"_2"O")/(1 cancel("mol H"_2"O")) = 7.72 "g H"_2"O"