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Answer :
There are 42/44 moles of CO2, 99.9/56 moles of KOH.
Since 42/44 moles of CO2 needs 84/44 moles of KOH to react with,
there aren't enough KOH in this case. Thus, (D) is the limiting reactant.
Since 42/44 moles of CO2 needs 84/44 moles of KOH to react with,
there aren't enough KOH in this case. Thus, (D) is the limiting reactant.
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Answer: The correct answer is Option D.
Explanation:
Limiting reagent is defined as the reagent which limits the formation of product and is present in less quantity as a reactant.
Excess reagent is defined as the reagent which is present in excess as a reactant.
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
- For carbon dioxide:
Given mass of carbon dioxide = 42 g
Molar mass of carbon dioxide = 44 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of carbon dioxide}=\frac{42g}{44g/mol}=0.954mol[/tex]
- For KOH:
Given mass of KOH = 99.9 g
Molar mass of KOH = 56.1 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of KOH}=\frac{99.9g}{56.1g/mol}=1.78mol[/tex]
For the given chemical reaction:
[tex]CO_2+2KOH\rightarrow K_2CO_3+H_2O[/tex]
By Stoichiometry of the reaction:
2 mole of potassium hydroxide reacts with 1 mole of carbon dioxide.
So, 1.78 moles of KOH will react with = [tex]\frac{1}{2}\times 1.78=0.89mol[/tex] of carbon dioxide.
As, the required amount of carbon dioixde is less than the given amount. Thus, it is considered as an excess reagent.
Potassium hydroxide is considered as a limiting reagent becase it limits the formation of products.
Thus, the correct answer is option D.
