Question

By the reaction of carbon and oxygen, a mixture of $CO$ and $C{O}_2$ is obtained. What is the composition of the mixture by mass obtained when $20$ g of $O_2$ reacts with $12$ g of carbon?

• A. $CO$ : $C{O}_2$ $=21:11$
• B. $CO$ : $C{O}_2$ $=11:21$
• C. $CO$ : $C{O}_2$ $=21:22$
• D. $CO$ : $C{O}_2$ $=42:11$

Answer 1

The correct option is A $CO$ : $C{O}_2$ $=21:11$

The balanced chemical equations for the combustion reaction are as follows:
$2xC+x{O}_2\to 2xCO$
$yC+y{O}_2\to yC{O}_2$
Total number of moles of $C$ $=2x+y$
Total number of moles of oxygen = $x+y$
The molar masses of $C$ and oxygen are $12$ g/mol and $32$ g/mol respectively.
$12$ g of $C$ corresponds to $1$ mole and $20$ g of oxygen corresponds to $\frac{20}{32}=0.625$ moles.
Hence, $2x+y=1$......(1)
and $x+y=0.625$......(2)
Substracting equation (2) from (1), we get
$(2x+y)-(x+y)=1-0.625$
$x=0.375$......(3)
Substituting equation (3) in equation (2), we get
$0.375+y=0.625$
$y=0.250$
The number of moles of $CO$ formed are $2x=2\left(0.375\right)=0.75$
The number of moles of $C{O}_2$ formed are $y=0.250$
The molar masses of $CO$ and $C{O}_2$ are $28$ g/mol and $44$ g/mol respectively.
The mass of $CO$ formed is $0.75\times 28=21g$
The mass of $C{O}_2$ formed is $0.250\times 44=11g$.
Hence, the composition of the mixture is $CO:C{O}_2=21:11$.