Question

When Carbon disulphide is burnt in an excess of Oxygen gas, the ratio of moles of ${\mathrm{CO}}_2$ to ${\mathrm{SO}}_2$ gases produced is :

• A. 1:1
• B. 1:2
• C. 2:1
• D. 2:3

Answer 1

The correct option is B

1:2

1. When carbon disulfide burns in the presence of Oxygen(${\mathrm{O}}_2$), Sulfur dioxide(${\mathrm{SO}}_2$) and Carbon dioxide (${\mathrm{CO}}_2$)are produced according to the following equation:

$$\begin{gathered} {\mathrm{CS}}_2\left(\mathrm{s}\right)+3{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{CO}}_2\left(\mathrm{g}\right)+2{\mathrm{SO}}_2\left(\mathrm{g}\right) \\ (\mathrm{Carbon}\mathrm{disulphide})\left(\mathrm{Oxygen}\right)(\mathrm{Carbon}\mathrm{dioxide})(\mathrm{Sulpur}\mathrm{dioxide}) \end{gathered}$$.

2. Therefore, the ratio of moles of ${\mathrm{CO}}_2$ to ${\mathrm{SO}}_2$ gases produced is 1 : 2.

Hence, the option (b) is correct. i.e., the ratio of moles of ${\mathrm{CO}}_2$ to ${\mathrm{SO}}_2$ gases produced is 1 : 2.