Answer :
To determine which of the reactants is the limiting reactant in the given chemical reaction, we'll follow these steps:
1. Write the Balanced Chemical Equation:
The balanced equation provided is:
[tex]\[
\text{CO}_2 + 2 \text{KOH} \rightarrow \text{K}_2\text{CO}_3 + \text{H}_2\text{O}
\][/tex]
This tells us that 1 mole of CO[tex]\(_2\)[/tex] reacts with 2 moles of KOH.
2. Calculate the Moles of Each Reactant:
- CO[tex]\(_2\)[/tex] Calculation:
- The molar mass of CO[tex]\(_2\)[/tex] is 44.01 g/mol.
- Given mass of CO[tex]\(_2\)[/tex]: 42.0 g.
- Moles of CO[tex]\(_2\)[/tex] = [tex]\(\frac{42.0 \text{ g}}{44.01 \text{ g/mol}} \approx 0.954\)[/tex] moles.
- KOH Calculation:
- The molar mass of KOH is 56.11 g/mol.
- Given mass of KOH: 99.9 g.
- Moles of KOH = [tex]\(\frac{99.9 \text{ g}}{56.11 \text{ g/mol}} \approx 1.780\)[/tex] moles.
3. Determine the Limiting Reactant:
According to the balanced equation, 1 mole of CO[tex]\(_2\)[/tex] requires 2 moles of KOH to react completely.
- Calculate the moles of KOH needed for 0.954 moles of CO[tex]\(_2\)[/tex]:
[tex]\[
0.954 \text{ moles CO}_2 \times 2 \text{ moles KOH/mol CO}_2 = 1.908 \text{ moles KOH needed}
\][/tex]
- Compare the available moles of KOH (1.780 moles) with the needed moles (1.908 moles).
Since we have only 1.780 moles of KOH available, which is less than the 1.908 moles needed, KOH is the limiting reactant.
Therefore, in this reaction, KOH is the limiting reactant.
1. Write the Balanced Chemical Equation:
The balanced equation provided is:
[tex]\[
\text{CO}_2 + 2 \text{KOH} \rightarrow \text{K}_2\text{CO}_3 + \text{H}_2\text{O}
\][/tex]
This tells us that 1 mole of CO[tex]\(_2\)[/tex] reacts with 2 moles of KOH.
2. Calculate the Moles of Each Reactant:
- CO[tex]\(_2\)[/tex] Calculation:
- The molar mass of CO[tex]\(_2\)[/tex] is 44.01 g/mol.
- Given mass of CO[tex]\(_2\)[/tex]: 42.0 g.
- Moles of CO[tex]\(_2\)[/tex] = [tex]\(\frac{42.0 \text{ g}}{44.01 \text{ g/mol}} \approx 0.954\)[/tex] moles.
- KOH Calculation:
- The molar mass of KOH is 56.11 g/mol.
- Given mass of KOH: 99.9 g.
- Moles of KOH = [tex]\(\frac{99.9 \text{ g}}{56.11 \text{ g/mol}} \approx 1.780\)[/tex] moles.
3. Determine the Limiting Reactant:
According to the balanced equation, 1 mole of CO[tex]\(_2\)[/tex] requires 2 moles of KOH to react completely.
- Calculate the moles of KOH needed for 0.954 moles of CO[tex]\(_2\)[/tex]:
[tex]\[
0.954 \text{ moles CO}_2 \times 2 \text{ moles KOH/mol CO}_2 = 1.908 \text{ moles KOH needed}
\][/tex]
- Compare the available moles of KOH (1.780 moles) with the needed moles (1.908 moles).
Since we have only 1.780 moles of KOH available, which is less than the 1.908 moles needed, KOH is the limiting reactant.
Therefore, in this reaction, KOH is the limiting reactant.
