Question

40 ml gaseous mixture of CO, CH4 and Ne was exploded with 10 ml of oxygen. On cooling the gases occupied 36.5 ml. After treatment with KOH the volume reduced by 9 ml and again on treatment with alkaline pyrogallol the volume is further reduced. What is the percentage of CH4 in the original mixture.

FULL ANSWER PLZZ​

Answer 1

Let volume of CO, CH4 and Ne be x, y and z respectively.

$CO+\frac{1}{2}O_2\rightarrow CO_2$.....(1)

here it is clear that x volume of CO reacts with x/2 volume of O2 and produces sx volume of CO2

and $CH_4+2O_2\rightarrow CO_2+2H_2O(l)$.....(2)

here it is clear that y volume of CH4 reacts with 2y volume of O2 and produces y volume of CO2.

now remaining volume of O2 = 10 - x/2 - 2y

volume after reaction,

volume of CO2 [eq(1)] + volume of CO2 [eq(2)] + volume of oxygen remaining + volume of Ne = 36.5 ml

or, x + y + 10 - x/2 - 2y + z = 36.5 ml

or, x/2 - y + z = 26.5 ml ......(i)

given, volume of Co + volume of CH4 + volume of Ne = 40ml

or, x + y + z = 40ml.....(ii)

we know, KOH reduces CO2

so, volume of CO2 [eq(1) ] + volume of CO2 [eq(2)] = 9ml

or, x + y = 9ml.....(iii)

solving equations (i), (ii) and (iii)

z = 31 , x = 3 and y = 6

so, volume of CH4 = 6ml

% of CH4 in the original mixture = 6/40 × 100 = 15%.

Answer 2

Answer:

Explanation:

Suppose, the volume of CO, CH4 and Ne are x, y and z respectively. CO+ 1/2O2à CO2 (1). It is clear that x volume of CO reacts with x/2 volume of O2 and produces sx volume of the CO2 and CH4+2O2 à CO2+2H2 O (l).....(2). It is clear that y volume of CH4 reacts with 2y volume of O2 and produces y volume of CO2. Hence the remaining volume of O2 = 10 - x/2 - 2y volume after reaction, volume of CO2 [eq(1)] + volume of CO2 [eq(2)] + volume of oxygen remaining + volume of Ne = 36.5 ml or, x + y + 10 - x/2 - 2y + z = 36.5 ml or, x/2 - y + z = 26.5 ml ......(i)Given, volume of Co + volume of CH4 + volume of Ne = 40ml or, x + y + z = 40ml.....(ii)we know, KOH reduces CO2 .so, volume of CO2 [eq(1) ] + volume of CO2 [eq(2)] = 9ml or, x + y = 9ml.....(iii)solving equations (i), (ii) and (iii)z = 31, x = 3 and y = 6 So, volume of CH4 = 6ml. Therefore, % of CH4 in the original mixture = 6/40 × 100 = 15%.