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Hint: the coke is considered as only carbon thus has a formula as C. Coke is reacted with the carbon dioxide such that coke is reduced to carbon monoxide. The chemical reaction of coke that is carbon C with carbon dioxide gives two moles of carbon monoxide. The reaction may form a mixture of gases which includes $\text{ CO }$ and unreacted $\text{ C}{{\text{O}}_{\text{2}}}\text{ }$.
Complete step by step answer:
Coke is a grey and hard coloured porous fuel material. It has a high content of carbon element with few impurities. It is chiefly made out of carbon with a minor amount of hydrogen, nitrogen, sulphur, and oxygen. The coke is considered as only carbon thus has a formula as C.
The chemical reaction of coke that is carbon C with carbon dioxide gives two moles of carbon monoxide. The reaction is as depicted below,
$\text{ C}{{\text{O}}_{\text{2}}}\text{ + C (coke) }\to \text{ 2CO }$
According to the given problem, one litre of carbon dioxide reacts with one mole of carbon from the coke and generates 2 litres of carbon monoxide.
We have given that $\text{ 1}\text{.4 L }$ of the mixture of gases. The mixture is of carbon monoxide and untreated carbon dioxide.
Let's consider that x litres of carbon dioxide reacts with the carbon and give 2x litres of carbon monoxide. The total volume of the mixture would be equal to the sum of the volume of $\text{ C}{{\text{O}}_{\text{2}}}\text{ }$and $\text{ CO }$.Thus,
$\text{ Total volume of mixture will be = C}{{\text{O}}_{\text{2}}}\text{ }+\text{ CO }$
We have given that $\text{ 1}\text{.4 L }$of the mixture of gas thus the total volume of the mixture would be,
$\text{ Volume V(mixture) = }{{\text{V}}_{\text{(new volume of C}{{\text{O}}_{\text{2}}}\text{) }}}+\text{ }{{\text{V}}_{\text{CO }}}\text{ = 1}\text{.4 L }$
Thus the new volume of the carbon dioxide would be equal to the difference in the initial volume to the carbon dioxide reacted which would be given as,
$\text{ }{{\text{V}}_{\text{(new volume of C}{{\text{O}}_{\text{2}}}\text{) }}}=\text{V(initial volume of C}{{\text{O}}_{\text{2}}})-\text{V(reacted C}{{\text{O}}_{\text{2}}})\text{ = }\left( 1-\text{x} \right)\text{L }$
We know that the volume of carbon monoxide formed is two times the volume of carbon dioxide. Thus the volume of carbon dioxide can be calculated as,
$\begin{align}
& \text{ }{{\text{V}}_{\text{(new volume of C}{{\text{O}}_{\text{2}}}\text{) }}}+{{\text{V}}_{\text{(CO)}}}\text{ = 1}\text{.4 L } \\
& \Rightarrow (1-\text{x) + 2x }=\text{ 1}\text{.4 } \\
& \Rightarrow \text{ x = 0}\text{.4 L } \\
\end{align}$
Thus the volume of carbon dioxide reacting with the coke is $\text{ 0}\text{.4 L }$ .
Now we can determine the new volume of carbon dioxide as follows,
$\begin{align}
& \text{ V(new C}{{\text{O}}_{\text{2}}})\text{ = V(initial volume C}{{\text{O}}_{\text{2}}})-\text{V(reacted C}{{\text{O}}_{\text{2}}}) \\
& \Rightarrow \text{V(new C}{{\text{O}}_{\text{2}}})\text{ = 1}-\text{x } \\
\end{align}$
Let's substitute the value of x (that is the volume of carbon dioxide) we have,
$\text{ V(new C}{{\text{O}}_{\text{2}}})\text{ = 1}-\text{x }\Rightarrow \text{ 1}-0.4\text{ = 0}\text{.6 L of C}{{\text{O}}_{\text{2}}}\text{ }$
Now we know that $\text{ }{{\text{V}}_{\text{(CO)}}}=\text{ 2x }$,thus
The volume of carbon dioxide or the product formed is equal to
$\text{ }{{\text{V}}_{\text{(CO)}}}=\text{ 2x }=\text{ 2}\times \text{0}\text{.4 = 0}\text{.8 L of CO }$
Therefore the composition of the product obtained by reaction of one litre of carbon dioxide with coke is $\text{ 0}\text{.6 L }$ of carbon dioxide and $\text{ 0}\text{.8 L }$a litre of carbon monoxide (total volume remains as$\text{ 1}\text{.4 L }$)
Hence, (C) is the correct option.
Note: Note that coke and carbon monoxide is used as the reducing agent in the smelting process. The bottom of the furnace reducing agent is a carbon (as coke).it initially oxidized to carbon dioxide. This carbon dioxide then rises in the furnace and reacts with coke and is reduced as carbon monoxide. The reaction is endothermic in nature.
Complete step by step answer:
Coke is a grey and hard coloured porous fuel material. It has a high content of carbon element with few impurities. It is chiefly made out of carbon with a minor amount of hydrogen, nitrogen, sulphur, and oxygen. The coke is considered as only carbon thus has a formula as C.
The chemical reaction of coke that is carbon C with carbon dioxide gives two moles of carbon monoxide. The reaction is as depicted below,
$\text{ C}{{\text{O}}_{\text{2}}}\text{ + C (coke) }\to \text{ 2CO }$
According to the given problem, one litre of carbon dioxide reacts with one mole of carbon from the coke and generates 2 litres of carbon monoxide.
We have given that $\text{ 1}\text{.4 L }$ of the mixture of gases. The mixture is of carbon monoxide and untreated carbon dioxide.
Let's consider that x litres of carbon dioxide reacts with the carbon and give 2x litres of carbon monoxide. The total volume of the mixture would be equal to the sum of the volume of $\text{ C}{{\text{O}}_{\text{2}}}\text{ }$and $\text{ CO }$.Thus,
$\text{ Total volume of mixture will be = C}{{\text{O}}_{\text{2}}}\text{ }+\text{ CO }$
We have given that $\text{ 1}\text{.4 L }$of the mixture of gas thus the total volume of the mixture would be,
$\text{ Volume V(mixture) = }{{\text{V}}_{\text{(new volume of C}{{\text{O}}_{\text{2}}}\text{) }}}+\text{ }{{\text{V}}_{\text{CO }}}\text{ = 1}\text{.4 L }$
Thus the new volume of the carbon dioxide would be equal to the difference in the initial volume to the carbon dioxide reacted which would be given as,
$\text{ }{{\text{V}}_{\text{(new volume of C}{{\text{O}}_{\text{2}}}\text{) }}}=\text{V(initial volume of C}{{\text{O}}_{\text{2}}})-\text{V(reacted C}{{\text{O}}_{\text{2}}})\text{ = }\left( 1-\text{x} \right)\text{L }$
We know that the volume of carbon monoxide formed is two times the volume of carbon dioxide. Thus the volume of carbon dioxide can be calculated as,
$\begin{align}
& \text{ }{{\text{V}}_{\text{(new volume of C}{{\text{O}}_{\text{2}}}\text{) }}}+{{\text{V}}_{\text{(CO)}}}\text{ = 1}\text{.4 L } \\
& \Rightarrow (1-\text{x) + 2x }=\text{ 1}\text{.4 } \\
& \Rightarrow \text{ x = 0}\text{.4 L } \\
\end{align}$
Thus the volume of carbon dioxide reacting with the coke is $\text{ 0}\text{.4 L }$ .
Now we can determine the new volume of carbon dioxide as follows,
$\begin{align}
& \text{ V(new C}{{\text{O}}_{\text{2}}})\text{ = V(initial volume C}{{\text{O}}_{\text{2}}})-\text{V(reacted C}{{\text{O}}_{\text{2}}}) \\
& \Rightarrow \text{V(new C}{{\text{O}}_{\text{2}}})\text{ = 1}-\text{x } \\
\end{align}$
Let's substitute the value of x (that is the volume of carbon dioxide) we have,
$\text{ V(new C}{{\text{O}}_{\text{2}}})\text{ = 1}-\text{x }\Rightarrow \text{ 1}-0.4\text{ = 0}\text{.6 L of C}{{\text{O}}_{\text{2}}}\text{ }$
Now we know that $\text{ }{{\text{V}}_{\text{(CO)}}}=\text{ 2x }$,thus
The volume of carbon dioxide or the product formed is equal to
$\text{ }{{\text{V}}_{\text{(CO)}}}=\text{ 2x }=\text{ 2}\times \text{0}\text{.4 = 0}\text{.8 L of CO }$
Therefore the composition of the product obtained by reaction of one litre of carbon dioxide with coke is $\text{ 0}\text{.6 L }$ of carbon dioxide and $\text{ 0}\text{.8 L }$a litre of carbon monoxide (total volume remains as$\text{ 1}\text{.4 L }$)
Hence, (C) is the correct option.
Note: Note that coke and carbon monoxide is used as the reducing agent in the smelting process. The bottom of the furnace reducing agent is a carbon (as coke).it initially oxidized to carbon dioxide. This carbon dioxide then rises in the furnace and reacts with coke and is reduced as carbon monoxide. The reaction is endothermic in nature.
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