# Question #7152a

Dec 12, 2014

In order to solve these kinds of problems, one must always use the balanced chemical equation and the ideal gas law, $P V = n R T$.

The balanced equation given is

$2 {C}_{4} {H}_{10 \left(g\right)} + 13 {O}_{2 \left(g\right)} \to 8 C {O}_{2 \left(g\right)} + 10 {H}_{2} O \left(\left(l\right)\right)$

Notice that we have $2 : 8$ (or a $1 : 4$) mole ratio between ${C}_{4} {H}_{10}$ and $C {O}_{2}$; this means that for every mole of ${C}_{4} {H}_{10}$ used in the reaction, 4 moles of $C {O}_{2}$ will be produced.

Now, since we don't have a mass or a number of ${C}_{4} {H}_{10}$ to go by, let's assume we start with $10.0 g$ of butane. Knowing that butane's molar mass is $58 \frac{g}{m o l}$, we can determine the number of moles from

${n}_{b u \tan e} = \frac{m}{m o l a r m a s s} = \frac{10.0 g}{58 \frac{g}{m o l}} = 0.17$ moles.

We thus get ${n}_{C {O}_{2}} = 4 \cdot {n}_{b u t a n e} = 0.68$ moles.

So, the volume produced in this case is

$V = \frac{n R T}{P} = \frac{0.68 \cdot 0.082 \cdot \left(273.15 + 23\right)}{1.00} = 16.5 L$

This method can be used for any mass of butane given...