50 ml of gaseous mixture of acetylene and ethylene is taken in a ratio of a : b requires 700 ml of air containing 20% by volume O 2 for complete combusion. Calculate the volume of air required for complete combution of a mixture (50 ml) having ratio b : a. (Report your answer divide by 25).
Text Solution
Verified by Experts27
(27)
Let volume of C 2 H 2 = x ml (since gases are assume ideal, therefore moles ∝ volume)
∴ volue of C 2 H 4 = 50 – x
C 2 H 2 + 
O 2 
2CO 2 + H 2 O
x 
x
C 2 H 4 + 3O 2 
2CO 2 + 2H 2 O
50 – x 3(50 – x)
Total volume of O 2 = 
x + 3(50 – x) = 700 × 
⇒ x = 20 ml
∴ volume of C 2 H 2 = 20 ml
∴ volume of C 2 H 4 = 30 ml
In new conditions
∴ volume of C 2 H 2 = 30 ml
∴ volume of C 2 H 4 = 20 ml
C 2 H 2 + 
O 2 
2CO 2 + H 2 O
30 
x
C 2 H 4 + 3O 2 
2CO 2 + 2H 2 O
20 3(20) ml
Total volume of O 2 = 
+ 3(20) = 135
∴ Total volume of air = 
= 675 ml
V = 675/25 = 27
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