# How do you determine the mass of carbon dioxide produced when 0.85 g of butane reacts with oxygen according to the following equation: 2C_4H_10 + 13O_2 -> 8CO_2 10H_2O?

Sep 6, 2016

${C}_{4} {H}_{10} \left(g\right) + \frac{13}{2} {O}_{2} \left(g\right) \rightarrow 4 C {O}_{2} \left(g\right) + 5 {H}_{2} O \left(l\right)$

$5.58 \times {10}^{-} 2 \text{moles of "CO_2" are produced} .$

#### Explanation:

$\text{Moles of butane}$ $=$ $\frac{0.85 \cdot g}{58.12 \cdot g \cdot m o {l}^{-} 1}$ $=$ $1.46 \times {10}^{-} 2 m o l$.

Given the stoichiometry, each moles of butane gives $4 \cdot m o l$ $C {O}_{2}$ upon complete combustion.

And thus $\text{moles of } C {O}_{2}$ $=$ $4 \times 1.46 \times {10}^{-} 2 m o l = 5.58 \times {10}^{-} 2 m o l$.

This is an equivalent mass of $5.58 \times {10}^{-} 2 m o l \times 44.01 \cdot g \cdot m o {l}^{-} 1 \cong 2.5 \cdot g$.

In my stoichiometric equation, I used an half equiv of dioxygen in order to balance the equation. I do this because it makes the 'rithmetic a bit easier.