Question

10ml of the gaseous mixture of CO H2 and NH3 are completely oxidised by 8 ml O2 if the original mixture of CO H2 and NH3 contains equal volume of CO and H2 then what is the volume % of NH3 in original mixture?

Answer 1

Answer on Question #80263 - Chemistry - Physical Chemistry

Question:

10ml of the gaseous mixture of CO H2 and NH3 are completely oxidised by 8 ml O2 if the original mixture of CO H2 and NH3 contains equal volume of CO and H2 then what is the volume % of NH3 in original mixture?

Solution:

$2 \mathrm{CO} + \mathrm{O}_{2} = 2 \mathrm{CO}_{2}$

$2 \mathrm{H}_{2} + \mathrm{O}_{2} = 2 \mathrm{H}_{2} \mathrm{O}$

$4 \mathrm{NH}_{3} + 7 \mathrm{O}_{2} = 4 \mathrm{NO}_{2} + 6 \mathrm{H}_{2} \mathrm{O}$

Let $V(CO) = V(H2) = xL$ , then $V(NH3) = (0.01 - 2x)L$ ;

So, $n(CO) = n(H2) = x / 22.4 \, \text{mol}$ , and $n(NH3) = (0.01 - 2x) / 22.4 \, \text{mol}$ ;

$n1(O2) = n2(O2) = n / 2(CO;H2) = x / 11.2 \, \text{mol}$ ;

$n3(O2) = 7 / 4n(NH3) = 7 / 4(0.01 - 2x) / 22.4;$

Vtotal(O2) = V1+V2+V3 = 2x+2x+(7/4 (0.01-2x)) = 0.01 L;

$0.01 = 4x + 0.0175 - 3.5x;$

$0.5x = 0.0075$

$X = 3.5 \, \text{mL}$