Mole ConceptMedium
Question
When $1{dm}^{3}$ of ${CO}_{2}$ gas is passed over hot coke the volume of gaseous mixture after complete reaction at STP becomes $1.4{dm}^{3}$. The composition of the gaseous mixture at STP is:
Options
A.$0.8{dm}^{3}$ of $CO,0.8{dm}^{3}$ of ${CO}_{2}$
B.$0.8{dm}^{3}$ of $CO,0.6{dm}^{3}$ of ${CO}_{2}$
C.$0.6{dm}^{3}$ of $CO,0.8{dm}^{3}$ of ${CO}_{2}$
D.$0.6{dm}^{3}$ of $CO,0.4{dm}^{3}$ of ${CO}_{2}$
Solution
Sol. [Key Concept: CO2 + hot coke → 2CO (Boudouard reaction)]
$$\begin{matrix} & {CO}_{2} + C(\text{ }s) \rightarrow 2CO \\ & 1 - x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2x \end{matrix}$$
$$\begin{matrix} & 1 - x + 2x = 1 + x = 1.4 \\ & x = 0.4{dm}^{3} \end{matrix}$$
Volume of ${CO}_{2} = 1 - 0.4 = 0.6{dm}^{3}$
Volume of $CO = 2 \times 0.4 = 0.8{dm}^{3}$
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