
Given that; ${\text{2500c}}{{\text{m}}^{\text{2}}}$ of oxygen $\left[ {{{\text{O}}_{\text{2}}}} \right]$ was burnt with ${\text{600c}}{{\text{m}}^{\text{2}}}$ of ethane $\left[ {{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}} \right]$. The volume of unused oxygen is.
A) ${\text{400cc}}$
B) ${\text{1200cc}}$
C) ${\text{0cc}}$
D) ${\text{1800cc}}$
Answer
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Hint: Ethane can react with Oxygen in two ways. If it forms Carbon dioxide, this reaction is called as complete combustion and if it forms Carbon monoxide and then the reaction is called as partial combustion. In both the cases, the oxygen source of combustion is air. Air is 29% oxygen by mol and 79% Nitrogen.
Complete step by step answer:
The chemical reaction here is:
Ethane + Oxygen $ \to $ Carbon dioxide + Water vapour
This reaction shows the complete combustion of ethane which is true for all hydrocarbons.
Complete combustion requires more Oxygen than the partial combustion. So, by introducing more Oxygen to the reactor we can preferentially favor the complete combustion reaction.
${\text{2}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}{\text{ + 7}}{{\text{O}}_{\text{2}}} \to {\text{4C}}{{\text{O}}_{\text{2}}}{\text{ + 6}}{{\text{H}}_{\text{2}}}{\text{O}}$ is the balanced equation.
${\text{2}}$ volumes of ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}$ needs ${\text{7}}$ volumes of ${{\text{O}}_{\text{2}}}$
Therefore, ${\text{600cc}}$ of ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}$ needs $\dfrac{{\text{7}}}{{\text{2}}}{{ \times 600 = 2100cc}}$ of ${{\text{O}}_{\text{2}}}$
Hence, the volume of unused ${{\text{O}}_{\text{2}}}{\text{ = 2500 - 2100 = 400cc}}$.
So, the correct answer is Option A.
Note: Oxygen is generally fed to the reaction in excess for several reasons. Here. Oxygen is fed in excess for combustion and the reason is not to generate products of interest but to harness the energy that’s released in these exothermic reactions for some purpose.
The negative point in adding the excess air is that all of the nitrogen which is inert in this case, makes its way to the flu gas, so we’re essentially increasing the amount of nitrogen that is in the flu gas. As we introduce more inert gas into the reactor and once, we consume all the fuel there’s more stuff in the reactor that absorbs the heat. So we are actually transferring less heat out of the reactor. So, there’s an optimization to increase the conversion of the fuel and maximize the heat transfer across the reactor.
Complete step by step answer:
The chemical reaction here is:
Ethane + Oxygen $ \to $ Carbon dioxide + Water vapour
This reaction shows the complete combustion of ethane which is true for all hydrocarbons.
Complete combustion requires more Oxygen than the partial combustion. So, by introducing more Oxygen to the reactor we can preferentially favor the complete combustion reaction.
${\text{2}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}{\text{ + 7}}{{\text{O}}_{\text{2}}} \to {\text{4C}}{{\text{O}}_{\text{2}}}{\text{ + 6}}{{\text{H}}_{\text{2}}}{\text{O}}$ is the balanced equation.
${\text{2}}$ volumes of ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}$ needs ${\text{7}}$ volumes of ${{\text{O}}_{\text{2}}}$
Therefore, ${\text{600cc}}$ of ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}$ needs $\dfrac{{\text{7}}}{{\text{2}}}{{ \times 600 = 2100cc}}$ of ${{\text{O}}_{\text{2}}}$
Hence, the volume of unused ${{\text{O}}_{\text{2}}}{\text{ = 2500 - 2100 = 400cc}}$.
So, the correct answer is Option A.
Note: Oxygen is generally fed to the reaction in excess for several reasons. Here. Oxygen is fed in excess for combustion and the reason is not to generate products of interest but to harness the energy that’s released in these exothermic reactions for some purpose.
The negative point in adding the excess air is that all of the nitrogen which is inert in this case, makes its way to the flu gas, so we’re essentially increasing the amount of nitrogen that is in the flu gas. As we introduce more inert gas into the reactor and once, we consume all the fuel there’s more stuff in the reactor that absorbs the heat. So we are actually transferring less heat out of the reactor. So, there’s an optimization to increase the conversion of the fuel and maximize the heat transfer across the reactor.
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