Answer :
We start with the balanced chemical equation
[tex]$$
\text{Br}_2 + 2\,\text{KI} \longrightarrow \text{I}_2 + 2\,\text{KBr}.
$$[/tex]
Step 1. Calculate the moles of each reactant.
For bromine ([tex]$\text{Br}_2$[/tex]), the molar mass is approximately [tex]$159.8 \, \text{g/mol}$[/tex]. Given [tex]$108 \, \text{g}$[/tex] of [tex]$\text{Br}_2$[/tex], the moles of [tex]$\text{Br}_2$[/tex] are
[tex]$$
\text{moles of } \text{Br}_2 = \frac{108 \, \text{g}}{159.8 \, \text{g/mol}} \approx 0.6758 \, \text{mol}.
$$[/tex]
For potassium iodide ([tex]$\text{KI}$[/tex]), the molar mass is approximately [tex]$166.0 \, \text{g/mol}$[/tex]. Given [tex]$98.2 \, \text{g}$[/tex] of [tex]$\text{KI}$[/tex], the moles of [tex]$\text{KI}$[/tex] are
[tex]$$
\text{moles of } \text{KI} = \frac{98.2 \, \text{g}}{166.0 \, \text{g/mol}} \approx 0.5916 \, \text{mol}.
$$[/tex]
Step 2. Identify the limiting reagent.
The balanced equation shows that [tex]$1 \, \text{mol}$[/tex] of [tex]$\text{Br}_2$[/tex] requires [tex]$2 \, \text{mol}$[/tex] of [tex]$\text{KI}$[/tex]. Therefore, for the available [tex]$\text{Br}_2$[/tex]:
[tex]$$
\text{Required moles of } \text{KI} = 0.6758 \, \text{mol} \times 2 \approx 1.3516 \, \text{mol}.
$$[/tex]
Since only [tex]$0.5916 \, \text{mol}$[/tex] of [tex]$\text{KI}$[/tex] is available, [tex]$\text{KI}$[/tex] is the limiting reagent.
Step 3. Determine the moles of [tex]$\text{I}_2$[/tex] produced.
The reaction shows that [tex]$2 \, \text{mol}$[/tex] of [tex]$\text{KI}$[/tex] produce [tex]$1 \, \text{mol}$[/tex] of [tex]$\text{I}_2$[/tex]. Therefore, the moles of [tex]$\text{I}_2$[/tex] formed are
[tex]$$
\text{moles of } \text{I}_2 = \frac{0.5916 \, \text{mol}}{2} \approx 0.2958 \, \text{mol}.
$$[/tex]
Step 4. Calculate the theoretical yield of [tex]$\text{I}_2$[/tex].
The molar mass of [tex]$\text{I}_2$[/tex] is approximately [tex]$253.8 \, \text{g/mol}$[/tex]. Thus, the theoretical mass of [tex]$\text{I}_2$[/tex] is
[tex]$$
\text{theoretical yield} = 0.2958 \, \text{mol} \times 253.8 \, \text{g/mol} \approx 75.07 \, \text{g}.
$$[/tex]
Rounding the result, the theoretical yield of [tex]$\text{I}_2$[/tex] is approximately [tex]$75.1 \, \text{g}$[/tex].
The correct answer is therefore [tex]$\boxed{75.1 \, \text{g}}$[/tex].
[tex]$$
\text{Br}_2 + 2\,\text{KI} \longrightarrow \text{I}_2 + 2\,\text{KBr}.
$$[/tex]
Step 1. Calculate the moles of each reactant.
For bromine ([tex]$\text{Br}_2$[/tex]), the molar mass is approximately [tex]$159.8 \, \text{g/mol}$[/tex]. Given [tex]$108 \, \text{g}$[/tex] of [tex]$\text{Br}_2$[/tex], the moles of [tex]$\text{Br}_2$[/tex] are
[tex]$$
\text{moles of } \text{Br}_2 = \frac{108 \, \text{g}}{159.8 \, \text{g/mol}} \approx 0.6758 \, \text{mol}.
$$[/tex]
For potassium iodide ([tex]$\text{KI}$[/tex]), the molar mass is approximately [tex]$166.0 \, \text{g/mol}$[/tex]. Given [tex]$98.2 \, \text{g}$[/tex] of [tex]$\text{KI}$[/tex], the moles of [tex]$\text{KI}$[/tex] are
[tex]$$
\text{moles of } \text{KI} = \frac{98.2 \, \text{g}}{166.0 \, \text{g/mol}} \approx 0.5916 \, \text{mol}.
$$[/tex]
Step 2. Identify the limiting reagent.
The balanced equation shows that [tex]$1 \, \text{mol}$[/tex] of [tex]$\text{Br}_2$[/tex] requires [tex]$2 \, \text{mol}$[/tex] of [tex]$\text{KI}$[/tex]. Therefore, for the available [tex]$\text{Br}_2$[/tex]:
[tex]$$
\text{Required moles of } \text{KI} = 0.6758 \, \text{mol} \times 2 \approx 1.3516 \, \text{mol}.
$$[/tex]
Since only [tex]$0.5916 \, \text{mol}$[/tex] of [tex]$\text{KI}$[/tex] is available, [tex]$\text{KI}$[/tex] is the limiting reagent.
Step 3. Determine the moles of [tex]$\text{I}_2$[/tex] produced.
The reaction shows that [tex]$2 \, \text{mol}$[/tex] of [tex]$\text{KI}$[/tex] produce [tex]$1 \, \text{mol}$[/tex] of [tex]$\text{I}_2$[/tex]. Therefore, the moles of [tex]$\text{I}_2$[/tex] formed are
[tex]$$
\text{moles of } \text{I}_2 = \frac{0.5916 \, \text{mol}}{2} \approx 0.2958 \, \text{mol}.
$$[/tex]
Step 4. Calculate the theoretical yield of [tex]$\text{I}_2$[/tex].
The molar mass of [tex]$\text{I}_2$[/tex] is approximately [tex]$253.8 \, \text{g/mol}$[/tex]. Thus, the theoretical mass of [tex]$\text{I}_2$[/tex] is
[tex]$$
\text{theoretical yield} = 0.2958 \, \text{mol} \times 253.8 \, \text{g/mol} \approx 75.07 \, \text{g}.
$$[/tex]
Rounding the result, the theoretical yield of [tex]$\text{I}_2$[/tex] is approximately [tex]$75.1 \, \text{g}$[/tex].
The correct answer is therefore [tex]$\boxed{75.1 \, \text{g}}$[/tex].
