Question

10ml of the gaseous mixture of co,h2 and nh3 are completely oxidised by 8ml O2.if 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

The percentage volume of NH₃ in original mixture is 3.5 mL.

Explanation:

The oxidation of hydrogen:

2 H₂ + O₂ → 2 H₂O

The oxidation of carbon monoxide:

2 CO + O₂ → 2 CO₂

The oxidation of ammonia:

4 NH₃ + 7 O₂ → 4 NO₂ + 6 H₂O

Now, we can consider that

V(CO) = V(H₂) = x L

V(NH₃) = (0.01 - 2x) L

Now, the moles are:

n(CO) = n(H₂) = x/22.4 moles

n(NH₃) = (0.01 - 2x)/22.4 moles

n1(O₂) = n2(O₂) = n/2(CO;H₂) = x/11.2 moles

n3(O₂) = 7/4 × n(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 mL