Question

The % volume of in a gaseous mixture of ${C_4}{H_{10}}$, $C{H_4}$ and CO is 40. When 200 ml of the mixture is burnt in excess of ${O_2}$. Find the volume (in ml) of $C{O_2}$
A.220
B.340
C.440
D.560

Answer 1

The given sample is a mixture of gases which consist of ${C_4}{H_{10}}$, $C{H_4}$ and CO. Now each of these gases contribute to the physical properties of the mixture. They do not contribute to the chemical properties as a whole, because they are still unreactive with each other and form no by products.

Complete step by step answer:
The given gaseous mixture is made up of 3 gases namely, ${C_4}{H_{10}}$, $C{H_4}$ and CO. Now, the total volume of the gaseous mixture is said to be 200 ml. Out of this, constitutes about 40% of the total volume of the given gaseous mixture. Hence, to calculate the actual volume of ${C_4}{H_{10}}$:
Volume of ${C_4}{H_{10}}$ = 40 % of the entire gaseous mixture
= $\dfrac{{40}}{{100}} \times$volume of the entire gaseous mixture
= 0.4 $\times$200ml
= 80 ml
Now the remainder of the gaseous mixture has a volume of about (200-80) = 120 ml.
The remainder of the gaseous mixture includes the gases $C{H_4}$ and CO. For the ease of calculation, let us consider the volume of $C{H_4}$ to be ‘x’ ml. Then the volume of CO will be of whatever remains, i.e ‘120 - x’ ml.
The conditions of the reaction of this gaseous mixture is given that the entire mixture is burnt in the presence of excess of oxygen. Considering that one mole of each constituent molecule is used, the reactions of the constituent gases of the mixture with oxygen can be given by:
$\mathop {C{H_4}}\limits_{'x'ml} + 2{O_2} \to \mathop {C{0_2}}\limits_{'x'ml} + 2{H_2}O$
$\mathop {CO}\limits_{(120 - x)ml} + \dfrac{1}{2}{O_2} \to \mathop {C{O_2}}\limits_{(120 - x)ml}$
$\mathop {{C_4}{H_{10}}}\limits_{80ml} + \dfrac{{13}}{2}{O_2} \to \mathop {4C{O_2}}\limits_{4 \times 80 = 320ml} + 5{H_2}O$
Hence the total volume of carbon dioxide produced can be calculated as:
Total volume of $C{O_2}$= (x) + (120-x) + (320) = 120 + 320 = 440 ml

Hence, Option C is the correct option.

Note:
The excess oxygen that has been used in the burning the mixture is responsible for deriving carbon dioxide from each of the constituent gases. In case if the oxygen was supplied in a limited quantity, then all the constituents would not have been fully ignited and hence, the amount of carbon dioxide would be less.