Question

8 l mixture of methane and propane on complete combustion gives 10l CO 2 ; what will be volume of CH 4 in mixture:​

Answer 1

Answer:

Propane-

C

3

H

8

+5O

2

→3CO

2

+4H

2

O

3 liters of propane would produce 9 liters of 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.

Hence option B is correct

Answer 2

The volume of $CH_{4}$ in the mixture is 7 L

Given : 8 L mixture of methane and propane on complete combustion gives 10 L of  $CO_{2}$

To Find : The volume of $CH_{4}$ in the mixture

Solution : The volume of $CH_{4}$ in the mixture is 7 L

Let the volume of  $CH_{4}$ in the mixture is x L

Then the volume of propane in the mixture will be (8-x) L

(because total of 8 L mixture is given to us in the question)

Complete combustion reaction will be

$CH_{4} +2O_{2} - > CO_{2} +2H_{2}O$   equation 1)

Here 1 L of  $CH_{4}$ produces 1 L of  $CO_{2}$

then,  x L of  $CH_{4}$ produces x L of  $CO_{2}$

Now the combustion reaction for propane is

$C_{3} H_{8} +5O_{2} - > 3CO_{2} +4H_{2}O$   equation 2)

Here 1 L of propane produces 3 L of carbon dioxide gas

So 8-x L of  propane will produces 3 (8-x)  L of carbon dioxide gas

Now total volume of carbon dioxide produced is

x +3 (8-x) = 24 - 2x

But in question it is given that total volume of carbon dioxide produced is 10 L

So 24 - 2x = 10

14 = 2x

x = 7

So The volume of $CH_{4}$ in the mixture is 7 L

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