Question

One litre of ${\text{C}}{{\text{O}}_{\text{2}}}$ is passed through red-hot coke. The volume becomes ${\text{1}}{\text{.4 litre}}$ at the same temperature and pressure. The composition of products is:
A ) ${\text{0}}{\text{.8 litre of C}}{{\text{O}}_{\text{2}}}{\text{and 0}}{\text{.6 litre of CO}}$
B ) ${\text{0}}{\text{.7 litre of C}}{{\text{O}}_{\text{2}}}{\text{and 0}}{\text{.7 litre of CO}}$
C ) ${\text{0}}{\text{.6 litre of C}}{{\text{O}}_{\text{2}}}{\text{and 0}}{\text{.8 litre of CO}}$
D ) ${\text{0}}{\text{.4 litre of C}}{{\text{O}}_{\text{2}}}{\text{and 1}}{\text{.0 litre of CO}}$

Answer 1

Under identical conditions of temperature and pressure, the reaction stoichiometry in terms of number of moles is same as the reaction stoichiometry in terms of volume in litres. Thus, one litre of carbon dioxide gas gives two litres of carbon monoxide gas.

Complete step by step answer:
Carbon dioxide gas reacts with red hot coke to form carbon monoxide gas. Write a balanced chemical equation for the conversion of carbon dioxide gas to carbon monoxide gas.
${\text{C}}{{\text{O}}_{\text{2}}} + {\text{ C}}\left( {{\text{red hot}}} \right){\text{ }} \to {\text{2 CO}}$
As per the above balanced chemical equation, one mole of carbon dioxide gas gives two moles of carbon monoxide gas. Since volumes of gases are measured under identical conditions of temperature and pressure, the reaction stoichiometry in terms of number of moles is the same as the reaction stoichiometry in terms of volume in litres. Thus, one litre of carbon dioxide gas gives two litres of carbon monoxide gas.
Initially one litre of carbon dioxide gas is present. Volume of carbon monoxide gas present initially is zero litres.
Let x litres of carbon dioxide are consumed in the reaction.

| Carbon dioxide gas| Carbon monoxide gas
---|---|---
Initial volume (Litres)| 1| 0
Change in volume (Litres)| $- x$ | $\;2x$
Final volume (Litres)| $1 - x$| $\;2x$

Total final volume is the sum of the final volume of carbon dioxide gas and the final volume of carbon monoxide gas.
Total final volume is $\left( {1 - x} \right) + 2x = 1 + x$. But the total final volume is ${\text{1}}{\text{.4 litre}}$ .

$$1 + x = 1.4{\text{ L}} \\\ {\text{x = }}\left( {{\text{1}}{\text{.4}} - {\text{1}}} \right){\text{ L}} \\\ {\text{x = 0}}{\text{.4 L}} \\\$$

The final volume of carbon dioxide gas is

$$1 - x = \left( {{\text{1}} - {\text{0}}{\text{.4}}} \right){\text{ L}} \\\ {\text{ = 0}}{\text{.6 L}} \\\$$

The final volume of carbon monoxide is

$${\text{2x = 2}}\left( {{\text{0}}{\text{.4 L}}} \right) \\\ = {\text{0}}{\text{.8 L}} \\\$$

Hence, the composition is given by the option C ) ${\text{0}}{\text{.6 litre of C}}{{\text{O}}_{\text{2}}}{\text{and 0}}{\text{.8 litre of CO}}$.

Note: Do not forget to balance the chemical equation. If an unbalanced chemical equation is used, then error will be introduced during calculation. For example, in the unbalanced chemical equation, ${\text{C}}{{\text{O}}_{\text{2}}} + {\text{ C}}\left( {{\text{red hot}}} \right){\text{ }} \to {\text{ CO}}$ neither the number of carbon atoms is balanced, nor the number of oxygen atoms is balanced. If an unbalanced equation is used, then x L of carbon dioxide gas will give x L of carbon monoxide gas, which is not true.