Question

A mixture of 10 L of butane and propane was subjected to combustion and total volume of CO2 evolved was equal to 35 L. The volume of propane in litres in the mixture is
Answer:​

Answer 1

Answer:

CO

2

(3×3=9)

Butane-

C

4

H

1

0+65O

2

→4CO

2

+5H

2

O

3 liters of butane would produce 12 liters of CO

2

($$3\times 4 = 12$$)

Liters of propane = X

Liters of butane = Y

X+Y=3

3X+4Y=10

Solve 2 simultaneous equations to get-

Y = 1

2 mole propane and 1 mole butane.So the ratio is 2:1.

Explanation:

HERE IS YOUR ANSWER

PLEASE MARK ME BRAINLIEST