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Answer :
To determine the limiting reactant in the reaction between carbon dioxide ([tex]\(CO_2\)[/tex]) and potassium hydroxide (KOH), follow these steps:
1. Determine the Moles of Each Reactant:
- The molar mass of [tex]\(CO_2\)[/tex] is approximately 44.01 g/mol.
- The molar mass of KOH is approximately 56.11 g/mol.
For [tex]\(CO_2\)[/tex]:
[tex]\[
\text{Moles of } CO_2 = \frac{42.0 \, \text{g}}{44.01 \, \text{g/mol}} = 0.954 \, \text{mol}
\][/tex]
For KOH:
[tex]\[
\text{Moles of KOH} = \frac{99.9 \, \text{g}}{56.11 \, \text{g/mol}} = 1.780 \, \text{mol}
\][/tex]
2. Use the Balanced Chemical Equation:
The balanced chemical equation is:
[tex]\[
CO_2 + 2 \, \text{KOH} \rightarrow K_2CO_3 + H_2O
\][/tex]
This tells us that 1 mole of [tex]\(CO_2\)[/tex] reacts with 2 moles of KOH.
3. Calculate the Required Moles of KOH for the Available [tex]\(CO_2\)[/tex]:
If you have 0.954 moles of [tex]\(CO_2\)[/tex], the required moles of KOH would be:
[tex]\[
\text{Required moles of KOH} = 0.954 \times 2 = 1.909 \, \text{mol}
\][/tex]
4. Identify the Limiting Reactant:
You have 1.780 moles of KOH available but need 1.909 moles for the complete reaction with the [tex]\(CO_2\)[/tex] present. Since you have fewer moles of KOH than needed, KOH is the limiting reactant.
In summary, even though you might have started with a considerable amount of both reactants, KOH runs out first and thus is the limiting reactant in this chemical reaction.
1. Determine the Moles of Each Reactant:
- The molar mass of [tex]\(CO_2\)[/tex] is approximately 44.01 g/mol.
- The molar mass of KOH is approximately 56.11 g/mol.
For [tex]\(CO_2\)[/tex]:
[tex]\[
\text{Moles of } CO_2 = \frac{42.0 \, \text{g}}{44.01 \, \text{g/mol}} = 0.954 \, \text{mol}
\][/tex]
For KOH:
[tex]\[
\text{Moles of KOH} = \frac{99.9 \, \text{g}}{56.11 \, \text{g/mol}} = 1.780 \, \text{mol}
\][/tex]
2. Use the Balanced Chemical Equation:
The balanced chemical equation is:
[tex]\[
CO_2 + 2 \, \text{KOH} \rightarrow K_2CO_3 + H_2O
\][/tex]
This tells us that 1 mole of [tex]\(CO_2\)[/tex] reacts with 2 moles of KOH.
3. Calculate the Required Moles of KOH for the Available [tex]\(CO_2\)[/tex]:
If you have 0.954 moles of [tex]\(CO_2\)[/tex], the required moles of KOH would be:
[tex]\[
\text{Required moles of KOH} = 0.954 \times 2 = 1.909 \, \text{mol}
\][/tex]
4. Identify the Limiting Reactant:
You have 1.780 moles of KOH available but need 1.909 moles for the complete reaction with the [tex]\(CO_2\)[/tex] present. Since you have fewer moles of KOH than needed, KOH is the limiting reactant.
In summary, even though you might have started with a considerable amount of both reactants, KOH runs out first and thus is the limiting reactant in this chemical reaction.
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