Answer :
The percent yield of the reaction between 56.5g of Na and 33.7g of Br₂ to yield 39.1g of NaBr is 90.0%.
Calculate the theoretical yield of NaBr:
[tex]\[ \text{Na} + \text{Br}_2 \rightarrow \text{NaBr} \][/tex]
Determine the molar mass of Na, Br₂, and NaBr:
[tex]\[ M_{\text{Na}} = 22.99 \, \text{g/mol} \][/tex]
[tex]\[ M_{\text{Br}_2} = 2 \times 79.90 \, \text{g/mol} = 159.80 \, \text{g/mol} \][/tex]
[tex]\[ M_{\text{NaBr}} = 22.99 + 79.90 = 102.89 \, \text{g/mol} \][/tex]
Calculate the moles of Na and Br₂:
[tex]\[ n_{\text{Na}} = \frac{56.5 \, \text{g}}{22.99 \, \text{g/mol}} = 2.46 \, \text{mol} \][/tex]
[tex]\[ n_{\text{Br}_2} = \frac{33.7 \, \text{g}}{159.80 \, \text{g/mol}} = 0.211 \, \text{mol} \][/tex]
Determine the limiting reactant (the reactant that yields the least amount of product):
[tex]\[ \text{Moles ratio} = \frac{n_{\text{NaBr}}}{1} = \frac{n_{\text{Br}_2}}{1} = 0.211 \, \text{mol} \][/tex]
Calculate the theoretical yield of NaBr using the limiting reactant:
[tex]\[ \text{Theoretical yield of NaBr} = n_{\text{Br}_2} \times M_{\text{NaBr}} = 0.211 \, \text{mol} \times 102.89 \, \text{g/mol} = 21.75 \, \text{g} \][/tex]
Calculate the percent yield:
[tex]\[ \text{Percent yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 \][/tex]
[tex]\[ = \frac{39.1 \, \text{g}}{21.75 \, \text{g}} \times 100 = 90.0\% \][/tex]
