Answer :
To determine the limiting reactant in the given chemical reaction, we need to compare the amount of moles we have for each reactant and see which one runs out first based on the stoichiometry of the reaction. The reaction given is:
[tex]\[ \text{CO}_2 + 2 \text{KOH} \rightarrow \text{K}_2\text{CO}_3 + \text{H}_2\text{O} \][/tex]
Here are the steps to solve the problem:
1. Determine the Molar Mass of Each Reactant:
- Molar mass of [tex]\( \text{CO}_2 \)[/tex] (carbon dioxide) is approximately 44.01 g/mol.
- Molar mass of [tex]\( \text{KOH} \)[/tex] (potassium hydroxide) is approximately 56.11 g/mol.
2. Calculate the Moles of Each Reactant:
- For [tex]\( \text{CO}_2 \)[/tex]:
[tex]\[
\text{Moles of } \text{CO}_2 = \frac{42.0 \, \text{g}}{44.01 \, \text{g/mol}} \approx 0.954 \, \text{mol}
\][/tex]
- For [tex]\( \text{KOH} \)[/tex]:
[tex]\[
\text{Moles of } \text{KOH} = \frac{99.9 \, \text{g}}{56.11 \, \text{g/mol}} \approx 1.780 \, \text{mol}
\][/tex]
3. Use Stoichiometry to Compare the Reactants:
- According to the balanced equation, 1 mole of [tex]\( \text{CO}_2 \)[/tex] reacts with 2 moles of [tex]\( \text{KOH} \)[/tex].
- Calculate the moles of [tex]\( \text{KOH} \)[/tex] needed for the available [tex]\( \text{CO}_2 \)[/tex]:
[tex]\[
\text{Moles of } \text{KOH needed} = 0.954 \, \text{mol} \times 2 = 1.908 \, \text{mol}
\][/tex]
4. Determine the Limiting Reactant:
- We have 1.780 mol of [tex]\( \text{KOH} \)[/tex] available, but we need 1.908 mol for the reaction.
- Since 1.780 mol is less than 1.908 mol, [tex]\( \text{KOH} \)[/tex] is the limiting reactant, as it will run out before all of the [tex]\( \text{CO}_2 \)[/tex] can react.
Thus, in this reaction, potassium hydroxide ([tex]\( \text{KOH} \)[/tex]) is the limiting reactant.
[tex]\[ \text{CO}_2 + 2 \text{KOH} \rightarrow \text{K}_2\text{CO}_3 + \text{H}_2\text{O} \][/tex]
Here are the steps to solve the problem:
1. Determine the Molar Mass of Each Reactant:
- Molar mass of [tex]\( \text{CO}_2 \)[/tex] (carbon dioxide) is approximately 44.01 g/mol.
- Molar mass of [tex]\( \text{KOH} \)[/tex] (potassium hydroxide) is approximately 56.11 g/mol.
2. Calculate the Moles of Each Reactant:
- For [tex]\( \text{CO}_2 \)[/tex]:
[tex]\[
\text{Moles of } \text{CO}_2 = \frac{42.0 \, \text{g}}{44.01 \, \text{g/mol}} \approx 0.954 \, \text{mol}
\][/tex]
- For [tex]\( \text{KOH} \)[/tex]:
[tex]\[
\text{Moles of } \text{KOH} = \frac{99.9 \, \text{g}}{56.11 \, \text{g/mol}} \approx 1.780 \, \text{mol}
\][/tex]
3. Use Stoichiometry to Compare the Reactants:
- According to the balanced equation, 1 mole of [tex]\( \text{CO}_2 \)[/tex] reacts with 2 moles of [tex]\( \text{KOH} \)[/tex].
- Calculate the moles of [tex]\( \text{KOH} \)[/tex] needed for the available [tex]\( \text{CO}_2 \)[/tex]:
[tex]\[
\text{Moles of } \text{KOH needed} = 0.954 \, \text{mol} \times 2 = 1.908 \, \text{mol}
\][/tex]
4. Determine the Limiting Reactant:
- We have 1.780 mol of [tex]\( \text{KOH} \)[/tex] available, but we need 1.908 mol for the reaction.
- Since 1.780 mol is less than 1.908 mol, [tex]\( \text{KOH} \)[/tex] is the limiting reactant, as it will run out before all of the [tex]\( \text{CO}_2 \)[/tex] can react.
Thus, in this reaction, potassium hydroxide ([tex]\( \text{KOH} \)[/tex]) is the limiting reactant.
