Question

20 ml of a mixture of C2H6 and C3H8 gases in X : Y mole ratio, is burnt completely by which 52 ml CO2 gas is evolved. If 20 ml of a mixture of CH4 and C4H10 gases in Y : X mole ratio is burnt completely then what is the volume (in ml) of CO2 gas evolved?
A) 40 B) 44 C) 50 D) 55

Answer 1

The volume of CO2 prodiced when 12 mL of CH4 and 8 mL of C4H10 is burnt is 56 mL.

What is the equation of the reaction?

The equation of the combustion of C2H6, CH4, C4H10, and C3H8 gases are given below:

2 C2H6 + 7 O2 ---> 4 CO2 + 6 H20

C3H8 + 5 O2 ---> 3 CO2 + 4 H20

CH4 + 2 O2 ----> CO2 + 2 H2O

2 C4H10 + 13 O2 ---> 8 CO2 + 10 H2O

X moles of C2H6 will produce 2X moles of CO2

Y moles of C3H8 will produce 3Y moles of CO2

Thus;

X + Y = 20 mL

X =20 - Y ---(1)

2X + 3Y = 52 mL ---(2)

Solving for X by substituting in (2)

2 (20 - Y) + 3Y = 52

40 - 2Y + 3Y = 52

Y = 12 mL

Then X = 8 mL

Volume of CO2 produced when 12 mL of CH4 and 8 mL of C4H10 is burnt = (12 × 2 + 8 × 4) = 56 mL

Therefore, the volume of CO2 produced is 56 mL.

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