# What mass of water is associated with the complete combustion of a 9.80*g mass of butane?

Nov 6, 2016

Approx. $15 \cdot g$ of water are produced.

#### Explanation:

We need (i) a stoichiometrically balanced equation:

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

and (ii), the equivalent quantity of butane combusted:

$\frac{9.80 \cdot g}{58.12 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.169 \cdot m o l$.

The given equation CLEARLY shows that 5 equiv of water are procuced per equiv of butane.

Therefore, $\text{mass of water}$

$=$ $0.169 \cdot m o l \times 5 \times 18.01 \cdot g \cdot m o {l}^{-} 1 = 15.2 \cdot g$

What mass of oxygen gas was required for this reaction?