Question

1L of $C{O_2}$ is passed red hot coke. The volume becomes $1.4L$ . The composition of the product(s) is:
(A) $0.6LCO$
(B) $0.8LC{O_2}$
(C) $0.6LC{O_2}$ and $0.8LCO$
(D) $0.8LC{O_2}$ and $0.6LCO$

Answer 1

Hint: When gas $C{O_2}$ passes through red hot coke, the reduction occurs and $C{O_2}$ $C{O_2} + C = 2CO$ gets reduce to $CO$ .However the total volume of the gases increases. Since it is already $CO$ in the lowest oxidation form, It does not react with charcoal.

Complete step by step answer:
When air is passed over red-hot coke, carbon monoxide gas is formed.
Here, $C{O_2}$ gets reduced $CO$ when treated with red hot coke, this is an exothermic reaction.
Chemical equation representing the reaction is: - $C{O_2} + C = 2CO$ .
From the above equation, we can say that $1mole$ of $C{O_2}$ produces $2mole$ of $CO$ when treated with hot coke. Gas or product that is produced When a mixture of air and steam is passed over red-hot coke is called Producer gas.

Given:
Initial volume $ = 1L$
And Final volume $ = 1.4L$
Now, let us say in $1L$ initial volume, $xL$ is $CO$ and $(1 - x)L$ is $C{O_2}$ . But after reaction $(1 - x)$ of $C{O_2}$ will be totally consumed and will give $2xCO$ .
Therefore,
Final volume becomes, $(1 - x) + 2x = 1.4$
Let us find $x$ ,
 $(1 - x) + 2x = 1.4$
 $1 + x = 1.4$
 $x = 0.4L$
As,
\[CO = 2x\]
$CO = 2 \times 0.4$
\[CO = 0.8L\]
Volume of $C{O_2} = 1.4 - 0.8 = 0.6L$
Finally, the composition of the product(s) is: $CO = 0.8L$ and $C{O_2} = 0.6L$

Hence option (C) $0.6LC{O_2}$ and $0.8LCO$ is correct.

Note: When $C{O_2}$ (g) is treated with hot coke it partially gets reduced to $CO$ (g). Coke may be a grey, hard, and porous fuel with high carbon content and few impurities, made by heating coal or oil within the absence of air—a fractionation process. It is a crucial industrial product, used mainly in ore smelting, but also as a fuel in stoves and forges when pollution may be a concern. A mixture of carbon monoxide and hydrogen, made by passing steam over red-hot coke is called water coke.