# How many moles of CO_2 form when 58.0 g of butane, C_4H_10, burn in oxygen?

Dec 1, 2016

$\text{58.0 g butane}$ in oxygen will produce $\text{3.99 mol carbon dioxide}$.

#### Explanation:

$\text{2C"_4"H"_10" + 13O"_2}$$\rightarrow$$\text{8CO"_2" + 10H"_2"O}$

Determine the molar mass of butane.

Molar Mass of $\text{C"_4"H"_10} :$

(4xx12.011"g/mol")+(10xx1.008"g/mol")="58.124 g/mol"

Determine the Mole Ratios for Butane and Carbon dioxide

$\left(2 \text{mol C"_4"H"_10)/(8"mol CO"_2}\right)$ and $\left(8 {\text{mol CO"_2)/(2"mol C"_4"H}}_{10}\right)$

Divide the given mass of butane by its molar mass (multiply by its reciprocal). This will give the moles of butane. Multiply times the molar ratio that places carbon dioxide in the numerator. This will give the moles of carbon dioxide.

58.0cancel("g C"_4"H"_10)xx(1cancel("mol C"_4"H"_10))/(58.124cancel("g C"_4"H"_10))xx(8"mol CO"_2)/(2cancel("mol C"_4"H"_10))="3.99 mol CO"_2"