# A student combusts 95.70g of butane (C_4H_10) in the presence of 94.40g of oxygen. How do you write and balance the equation for the reaction?

Feb 24, 2016

For complete combustion:

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

#### Explanation:

$\text{Moles of butane}$ $=$ $\frac{95.7 \cdot g}{58.12 \cdot g \cdot m o {l}^{-} 1}$ $=$ ?? $\text{mol}$.

$\text{Moles of oxygen}$ $=$ $\frac{94.4 \cdot g}{32.00 \cdot g \cdot m o {l}^{-} 1}$ $=$ ?? $\text{mol}$.

It is pretty clear that there is insufficient oxygen for complete combustion to occur. Butane would combust incompletely in these circumstances to give elemental carbon (as soot) and carbon monoxide gas. We have no handle on the amounts of each.