# The combustion of butane, C_4H_10, produces carbon dioxide water. When one sample of butane was burned, 4.46 grams of water was formed. How many grams of oxygen gas were consumed?

Mar 3, 2016

${\text{10.3 g O}}_{2}$

#### Explanation:

Combustion reactions with hydrocarbons (compounds made up of only carbon and hydrogen, ${C}_{4} {H}_{10}$) involve the hydrocarbon reacting with ${O}_{2}$ gas to form ${H}_{2} O$ and $C {O}_{2}$

We must first balance the equation, we end up with;

$2 {C}_{4} {H}_{10} + 13 {O}_{2} \to 10 {H}_{2} O + 8 C {O}_{2}$

with the balanced equation, most of the work is done. We now simply convert from ${H}_{2} O$ to ${O}_{2}$ using the mol to mol ratios.

$H = \text{1.01 g}$
$O = \text{16.00 g}$
${H}_{2} O = \text{18.02 g}$
${O}_{2} = \text{32.00 g}$

$\text{4.46 g H"_2"O" xx("1 mol H"_2"O")/("18.02 g H"_2"O") xx("13 mol O"_2)/("10 mol H"_2"O") xx ("32.0 g O"_2)/("1 mol O"_2) = "10.3 g O"_2" gas}$