Question #7152a

Question

Question #7152a

1 Answer

Impact of this question

5024 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-12-12T11:27:18

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

The balanced equation given is

$2 C_{4} H_{10 (g)} + 13 O_{2 (g)} \to 8 C O_{2 (g)} + 10 H_{2} O ((l))$ $2C_4H_(10(g)) + 13O_(2(g)) -> 8CO_(2(g)) + 10H_2O((l))$

Notice that we have $2 : 8$ $2:8$ (or a $1 : 4$ $1:4$ ) mole ratio between $C_{4} H_{10}$ $C_4H_10$ and $C O_{2}$ $CO_2$ ; this means that for every mole of $C_{4} H_{10}$ $C_4H_10$ used in the reaction, 4 moles of $C O_{2}$ $CO_2$ will be produced.

Now, since we don't have a mass or a number of $C_{4} H_{10}$ $C_4H_10$ to go by, let's assume we start with $10.0 g$ $10.0g$ of butane. Knowing that butane's molar mass is $58 \frac{g}{m o l}$ $58g/(mol)$ , we can determine the number of moles from

$n_{b u tan e} = \frac{m}{m o l a r m a s s} = \frac{10.0 g}{58 \frac{g}{m o l}} = 0.17$ $n_(butane) = m/(molarmass) = (10.0g)/(58g/(mol)) = 0.17$ moles.

We thus get $n_{C O_{2}} = 4 \cdot n_{b u t a n e} = 0.68$ $n_(CO_2) = 4 * n_(bu t an e) = 0.68$ moles.

So, the volume produced in this case is

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

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

Science

Math

Humanities

... and beyond

Question #7152a

1 Answer

Related questions

Impact of this question