# What is the mass of water produced when 9.33 g of butane reacts with excess oxygen?

Mar 21, 2016

5xx0.161*molxx18.01*g*mol^-1=??

#### Explanation:

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

$\text{Moles of butane}$ $=$ $\frac{9.33 \cdot g}{58.12 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.161 \cdot m o l$.

We assume complete combustion, and know that each mole of butane gives off 5 mol of water upon combustion.

Given that we know the molar quantity of butane, we simply perform the operation, 5xx0.161*molxx18.01*g*mol^-1=??

What is the mass of carbon dioxide gas evolved in the reaction?