Question

20ml of a mixture of ${{C}_{2}}{{H}_{2}}$ and $CO$ was exploded with 30 ml of oxygen. The gases after the reaction had a volume of 34 ml. on treatment with KOH, 8 ml of oxygen remained. Calculate the composition of the mixture.
(a) ${{C}_{2}}{{H}_{2}} = 6ml,CO = 14ml$
(b) ${{C}_{2}}{{H}_{2}} = 4ml,CO = 28ml$
(c) ${{C}_{2}}{{H}_{2}} = 8ml,CO = 30ml$
(d) ${{C}_{2}}{{H}_{2}} = 10ml,CO = 19ml$

Answer 1

. We know the volume of the mixture before reacting and from this, we can assume the volume of ${{C}_{2}}{{H}_{2}}$ and $CO$. Now, we have been given the total volume of the mixture produced after the reaction, and the volume of oxygen remained and from this, we can find the volume of gases carbon dioxide produced after the reaction. So, thus, by comparing the volume of gases carbon dioxide produced after reaction with the volume that came after subtraction, we can easily find the composition of the reaction mixture. Now solve it.

Complete step by step answer:
The mixture consists of gases ethene and carbon monoxide which when exploded with oxygen gas results in the formation of carbon dioxide gas. The reactions occur as;

& {{C}_{2}}{{H}_{2}}+{{O}_{2}}\to 2C{{O}_{2}}+{{H}_{2}}O---------(1) \\\ & \\\ & \\\ & CO+{{O}_{2}}\to C{{O}_{2}}----------(2) \\\ \end{aligned}$$ The volume of total mixture = 20 ml (given) Then, the volume of ${{C}_{2}}{{H}_{2}}$ in the mixture in equation (1)= x ml And the volume of $C{{O}_{2}}$ in the mixture in the equation (1)=2x the volume of $CO$ in the mixture in the equation (2)= 20-x ml And the volume of $C{{O}_{2}}$ in the mixture in the equation (2)=20-x ml then, the equation (1) can be written as; Then after the reaction; The mixture of the volume =34 ml (given) And then, the volume of oxygen remained = 8 ml (given) Since, after the reaction, the gas present in the mixture is carbon dioxide along with some oxygen gas. Then , the volume of the gas carbon dioxide in the mixture = $$\Rightarrow 34-8\text{ }=26\text{ }ml$$---(3) Then, taking the concentration of carbon dioxide from equation (1) and (2) and comparing with the equation (3); we get: $\begin{aligned} & \Rightarrow 2x + 20 - x = 26 \\\ & \Rightarrow 20 + x = 26 \\\ & \Rightarrow x = 26 - 20 \\\ & \Rightarrow x = 6 \\\ \end{aligned}$ As, we know x is the volume of ${{C}_{2}}{{H}_{2}}$ in the mixture, so, volume of ${{C}_{2}}{{H}_{2}}$ = 6 ml the volume of $CO$ in the mixture=$$\Rightarrow 20-x\text{ }=20-6\text{ }=14\text{ }ml.$$. **So, the correct answer is “Option A”.** **Note:** While comparing the volume of gases with the resultant volume of the mixture after the reaction, don’t take the volume of ${{C}_{2}}{{H}_{2}}$ and $CO$ because after being exploded with the oxygen gas, these two gases results in the formation of carbon dioxide gas along with the water and mixture consists of carbon dioxide gas along with some oxygen gas.