Question

Percentage by volume of c3h8, in a gaseous mixture
of C3h8,CH4, and CO is 20. When 100 ml of the
mixture is burnt in excess of O2, the volume of CO2
produced at STP will be
(1) 90 ml
(2) 160 ml
(3) 140 ml
(4) 200 ml​

Answer 1

Answer : The correct option is, (3) 140 ml

Explanation :

As we are given:

Volume of $C_3H_8$ = 20 ml = 0.02 L     (1 L = 1000 ml)

Volume of mixture = 100 ml

Let the volume of $CH_4$ and $CO$ be, 'x' and 'y' respectively.

Volume of mixture = Volume of $C_3H_8$ + Volume of $CH_4$ + Volume of $CO$

100 = 20 + x + y

x + y = 80 ml = 0.08 L

The balanced chemical reactions are:

(1) $C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

(2) $CH_4+2O_2\rightarrow CO_2+2H_2O$

(3) $CO+\frac{1}{2}O_2\rightarrow CO_2$

From the 1st balanced chemical reaction we conclude that,

As, 1 L of $C_3H_8$ react to give 3 L of $CO_2$

As, 0.02 L of $C_3H_8$ react to give $0.02\times 3=0.06L$ of $CO_2$

From the 2nd balanced chemical reaction we conclude that,

As, 1 L of $CH_4$ react to give 1 L of $CO_2$

As, 'x' L of $CH_4$ react to give x L of $CO_2$

From the 3rd balanced chemical reaction we conclude that,

As, 1 L of $CO$ react to give 1 L of $CO_2$

As, 'y' L of $CH_4$ react to give y L of $CO_2$

As we know that,

Volume of $CO_2$ = Volume of $C_3H_8$ + Volume of $CH_4$ + Volume of $CO$

Volume of $CO_2$ = 0.06 + x + y

And,

x + y = 0.08 L

Volume of $CO_2$ = 0.06 + 0.08 = 0.14 L = 140 ml

Therefore, the volume of $CO_2$ produced at STP will be, 140 ml

Answer 2

Answer:

option 3

if question no is 40 then like my answer