# How many molecules of water will be produced if 52.6g of pentane are burned?

Jul 4, 2016

Approx. $3.5 \times {N}_{A}$, where ${N}_{A} = \text{Avogadro's Number}$

#### Explanation:

We need (i) a balanced chemical equation:

${C}_{5} {H}_{12} \left(l\right) + 8 {O}_{2} \left(g\right) \rightarrow 5 C {O}_{2} \left(g\right) + 6 {H}_{2} O \left(l\right)$

And thus for each mole of pentane combusted, (ii) 6 moles of water are produced.

$\text{Moles of pentane}$ $=$ $\frac{52.6 \cdot g}{72.15 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.729 \cdot m o l$

And thus $5 \times 0.729 \cdot m o l$ of water are produced after complete combustion. Since in $1$ $m o l$ there are $6.022 \times {10}^{23}$ individual particles, there are how many water molecules?