# When propane, C_3H_8, is burned, carbon dioxide and water vapor are produced according to the following reaction: C_3H_8 + 5O_2 -> 3CO_2 + 4H_2O. How much propane is burned if 160.0 g of O2 are used and 132 g of CO2 and 72.0 g of H2O are produced?

Dec 19, 2016

${C}_{3} {H}_{8} + 5 {O}_{2} \rightarrow 3 C {O}_{2} + 4 {H}_{2} O$

Precisely one mole, $44 \cdot g$ of pentane, were combusted.

#### Explanation:

$\text{Moles of carbon dioxide:}$ $\frac{132 \cdot g}{44.0 \cdot g \cdot m o {l}^{-} 1} = 3 \cdot m o l$

$\text{Moles of water:}$ $\frac{72 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1} = 4 \cdot m o l$

$\text{Moles of oxygen:}$ $\frac{160 \cdot g}{32.0 \cdot g \cdot m o {l}^{-} 1} = 5 \cdot m o l$

Because there were $3$ $m o l$ carbon dioxide produced, and stoichiometric water, one mole precisely of propane was combusted.

If pentane were burned in excess dioxygen, and $220 \cdot g$ carbon dioxide were produced, what was the starting mass of pentane?

${C}_{5} {H}_{12} + 8 {O}_{2} \rightarrow 5 C {O}_{2} + 6 {H}_{2} O$