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]

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Rewritten by : Barada

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