# Question #99fa3

Oct 1, 2017

$15.7 g$

#### Explanation:

With the reaction balanced, the mass balance (conservation of mass) means that the reactants must also total 56.4g. So first, let’s convert the molar ratio into a mass ratio.
${C}_{4} {H}_{10} O = 74 g$ $6 {O}_{2} = 192 g$

The total normalized mass is 266g. We have only 56.4g, so we only have $\frac{56.4}{266} = 0.212$ of a “unit”.
Applying that to the molar ratios, we obtain
$0.212 \times \left({C}_{4} {H}_{10} O + 6 {O}_{2}\right) = 0.212$ moles ${C}_{4} {H}_{10} O$ and $1.27$ moles of ${O}_{2}$.

$74 \frac{g}{m o l} {C}_{4} {H}_{10} O \times 0.212 = 15.7 g$
$32 \frac{g}{m o l} {O}_{2} \times 1.27 = 40.7 g$
CHECK: $15.7 + 40.7 = 56.4$ CORRECT!