Answer :
We begin by calculating the number of moles of the reactants.
First, for bromine, we have:
[tex]$$
\text{Moles of } Br_2 = \frac{108.0\ \text{g}}{159.8\ \text{g/mol}} \approx 0.6758\ \text{mol}.
$$[/tex]
Next, for potassium iodide:
[tex]$$
\text{Moles of } KI = \frac{98.2\ \text{g}}{166.0\ \text{g/mol}} \approx 0.5916\ \text{mol}.
$$[/tex]
The balanced chemical equation is:
[tex]$$
Br_2 + 2KI \rightarrow 2KBr + I_2.
$$[/tex]
According to the stoichiometry, 2 moles of [tex]$KI$[/tex] produce 1 mole of [tex]$I_2$[/tex]. Therefore, the theoretical moles of [tex]$I_2$[/tex] that can be formed from [tex]$KI$[/tex] are:
[tex]$$
\text{Theoretical moles of } I_2 = \frac{0.5916\ \text{mol}}{2} \approx 0.2958\ \text{mol}.
$$[/tex]
Next, we compute the theoretical mass of [tex]$I_2$[/tex] using its molar mass (approximately [tex]$253.8\ \text{g/mol}$[/tex]):
[tex]$$
\text{Theoretical mass of } I_2 = 0.2958\ \text{mol} \times 253.8\ \text{g/mol} \approx 75.07\ \text{g}.
$$[/tex]
The percent yield is then calculated using the actual yield of [tex]$I_2$[/tex], which is [tex]$72.1\ \text{g}$[/tex]:
[tex]$$
\text{Percent yield} = \left(\frac{72.1\ \text{g}}{75.07\ \text{g}}\right) \times 100\% \approx 96.0\%.
$$[/tex]
Thus, the percent yield of [tex]$I_2$[/tex] is [tex]$\boxed{96.0\%}$[/tex].
First, for bromine, we have:
[tex]$$
\text{Moles of } Br_2 = \frac{108.0\ \text{g}}{159.8\ \text{g/mol}} \approx 0.6758\ \text{mol}.
$$[/tex]
Next, for potassium iodide:
[tex]$$
\text{Moles of } KI = \frac{98.2\ \text{g}}{166.0\ \text{g/mol}} \approx 0.5916\ \text{mol}.
$$[/tex]
The balanced chemical equation is:
[tex]$$
Br_2 + 2KI \rightarrow 2KBr + I_2.
$$[/tex]
According to the stoichiometry, 2 moles of [tex]$KI$[/tex] produce 1 mole of [tex]$I_2$[/tex]. Therefore, the theoretical moles of [tex]$I_2$[/tex] that can be formed from [tex]$KI$[/tex] are:
[tex]$$
\text{Theoretical moles of } I_2 = \frac{0.5916\ \text{mol}}{2} \approx 0.2958\ \text{mol}.
$$[/tex]
Next, we compute the theoretical mass of [tex]$I_2$[/tex] using its molar mass (approximately [tex]$253.8\ \text{g/mol}$[/tex]):
[tex]$$
\text{Theoretical mass of } I_2 = 0.2958\ \text{mol} \times 253.8\ \text{g/mol} \approx 75.07\ \text{g}.
$$[/tex]
The percent yield is then calculated using the actual yield of [tex]$I_2$[/tex], which is [tex]$72.1\ \text{g}$[/tex]:
[tex]$$
\text{Percent yield} = \left(\frac{72.1\ \text{g}}{75.07\ \text{g}}\right) \times 100\% \approx 96.0\%.
$$[/tex]
Thus, the percent yield of [tex]$I_2$[/tex] is [tex]$\boxed{96.0\%}$[/tex].
