# In a butane lighter, 9.7 g of butane combine with with 34.7 g of oxygen to form g g carbon dioxide and 29.3 how many grams of water?

Jul 14, 2017

Approx. $15 \cdot g$......

#### Explanation:

We need a stoichiometric 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 $\text{moles of butane} = \frac{9.7 \cdot g}{58.12 \cdot g \cdot m o {l}^{-} 1} = 0.167 \cdot m o l$.

And so upon complete combustion we gets......

$0.167 \cdot m o l \times 5 \times 18.011 \cdot g \cdot m o {l}^{-} 1 = 15.0 \cdot g$ water.......

We could have avoided this rigmarole by noting that mass is always with conserved in a chemical reaction. By the problem's specification, we had $\left(34.7 + 9.7\right) \cdot g = 44.4 \cdot g$ of reactant; we got a mass $29.3 \cdot g$ carbon dioxide, and thus there was a balance of $44.4 - 29.3 \cdot g = 15.0 \cdot g$ as required........WHICH MUST REPRESENT THE MASS OF WATER...........