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Which is the limiting reactant in the following reaction, given that you start with 42.0 g of [tex]$CO_2$[/tex] and 99.9 g of KOH?

Reaction: [tex]$CO_2 + 2 \, KOH \rightarrow K_2CO_3 + H_2O$[/tex]

A. [tex]$H_2O$[/tex]
B. [tex]$CO_2$[/tex]
C. KOH
D. [tex]$K_2CO_3$[/tex]

Answer :

To determine the limiting reactant in the reaction between [tex]\( CO_2 \)[/tex] and [tex]\( KOH \)[/tex], follow these steps:

1. Identify the Molar Masses:
- Molar mass of [tex]\( CO_2 \)[/tex]: 44.01 g/mol
- Molar mass of [tex]\( KOH \)[/tex]: 56.11 g/mol

2. Calculate Moles of Each Reactant:
- For [tex]\( CO_2 \)[/tex]:
[tex]\[
\text{Moles of } CO_2 = \frac{\text{Mass of } CO_2}{\text{Molar mass of } CO_2} = \frac{42.0 \, \text{g}}{44.01 \, \text{g/mol}} \approx 0.954 \, \text{mol}
\][/tex]
- For [tex]\( KOH \)[/tex]:
[tex]\[
\text{Moles of } KOH = \frac{\text{Mass of } KOH}{\text{Molar mass of } KOH} = \frac{99.9 \, \text{g}}{56.11 \, \text{g/mol}} \approx 1.780 \, \text{mol}
\][/tex]

3. Use the Balanced Chemical Equation:
- The balanced reaction is:
[tex]\[
CO_2 + 2 \, KOH \rightarrow K_2CO_3 + H_2O
\][/tex]
- This indicates that 1 mole of [tex]\( CO_2 \)[/tex] reacts with 2 moles of [tex]\( KOH \)[/tex].

4. Determine Stoichiometric Requirement:
- Calculate how many moles of [tex]\( KOH \)[/tex] are needed to fully react with the available moles of [tex]\( CO_2 \)[/tex]:
[tex]\[
\text{Required moles of } KOH = \text{Moles of } CO_2 \times 2 = 0.954 \, \text{mol} \times 2 = 1.908 \, \text{mol}
\][/tex]

5. Identify the Limiting Reactant:
- Compare the available moles of [tex]\( KOH \)[/tex] to the required moles:
- Available moles of [tex]\( KOH \)[/tex]: 1.780 mol
- Required moles of [tex]\( KOH \)[/tex]: 1.908 mol

Since the available moles of [tex]\( KOH \)[/tex] (1.780 mol) is less than the required moles (1.908 mol), [tex]\( KOH \)[/tex] is the limiting reactant.

Therefore, the limiting reactant is [tex]\( KOH \)[/tex].

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