Question

Question #b0afa

Answer 1

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

Answer 2

1.67%

Explanation:

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

`2CH_4 + 3O_2 rarr 2CO_2 + 4H_2O`

Then, determined the amount of `CO_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

`=(9.6 g CH4) xx ((1 mol CH4)/(16.042g CH4)) xx ((2 mol CO2)/(2 mol CH4)) xx ((44.01g CO2)/(1 mol CO2))`

`= 26.34 g` `CO_2`

Then to calculate the percent purity, divide the actual yield of `CO_2` with the theoretical yield:

`percent` `purity`

`= ((0.44g)/(26.34g)) xx 100`

`= 1.67%`