# In the complete combustion of 454.0g of liquid butane (C_4H_10) what mass of water will be produced?

Dec 28, 2016

Approx. $700 \cdot g$ 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 thus 5 equiv of water are produced per equiv of butane.

And (ii) we need the equivalent quantity of gas burnt:

$\text{Moles of butane}$ $=$ $\frac{454 \cdot g}{58.12 \cdot g \cdot m o {l}^{-} 1} = 7.81 \cdot m o l$

And thus 7.81*molxx5xx18.01*g*mol^-1=??g