Question

When burnt in air, a $12.0$ g mixture of carbon and sulphur yields a mixture of $C{O}_2$ and $S{O}_2$ is produced, in which the number of moles of $S{O}_2$ is half that of $C{O}_2$. The mass of the carbon the mixture contains is: (At. wt. $S=32$)

• A. $4.08g$
• B. $5.14g$
• C. $8.74g$
• D. $1.54g$

Answer 1

The correct option is B $5.14g$

according to the question, the number of moles of sulphur dioxide is one half the number of moles of carbon dioxide.

Hence, the number of moles of sulphur will be one half the number of moles of carbon. Let the mixture contains $x$ g of carbon and $12-x$ g of sulphur

The no of moles of carbon and sulphur are $x/44$ and $(12-x)/64$ respectively.

According to the question, $\frac{x}{44}=2\left(\frac{12-x}{64}\right)$.

So, $x=5.14g$. So the correct answer is B