# Question b0afa

Feb 12, 2016

As per eqn: CH4 +2O2 ----> CO2 + 2H2O, 1 mole or 44g CO2 is produced from 16 g methane.So 0.44 g Co2 will be obtained from 0.16 g methane. So 9.6 g sample contains 0.16g methane.

#### Explanation:

Hence% of purity is 0.16X100/9.6 = 1.67

Feb 12, 2016

1.67%

#### Explanation:

To determine the percent purity, first determine the balanced chemical reaction for the combustion of methane:

$2 C {H}_{4} + 3 {O}_{2} \rightarrow 2 C {O}_{2} + 4 {H}_{2} O$

Then, determined the amount of $C {O}_{2}$ produced when 9.6 grams of the sample undergoes complete combustion (also known as the theoretical yield. Our main assumption here is that the 9.6 grams of the sample is pure methane.

theoretical yield

$= \left(9.6 g C H 4\right) \times \left(\frac{1 m o l C H 4}{16.042 g C H 4}\right) \times \left(\frac{2 m o l C O 2}{2 m o l C H 4}\right) \times \left(\frac{44.01 g C O 2}{1 m o l C O 2}\right)$

$= 26.34 g$ $C {O}_{2}$

Then to calculate the percent purity, divide the actual yield of $C {O}_{2}$ with the theoretical yield:

$p e r c e n t$ $p u r i t y$

$= \left(\frac{0.44 g}{26.34 g}\right) \times 100$

= 1.67%#