Question

Which is the limiting reactant in the following reaction given that you start with 42.0 g of CO2 and 99.9 g KOH?

Answer 1

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:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

• 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:

$\text{Moles of carbon dioxide}=\frac{42g}{44g/mol}=0.954mol$

• For KOH:

Given mass of KOH = 99.9 g

Molar mass of KOH = 56.1 g/mol  

Putting values in above equation, we get:  

$\text{Moles of KOH}=\frac{99.9g}{56.1g/mol}=1.78mol$

For the given chemical reaction:

$CO_2+2KOH\rightarrow K_2CO_3+H_2O$

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 = $\frac{1}{2}\times 1.78=0.89mol$ 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.

Answer 2

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.