Question

A chemist reacted 57.50 grams of sodium metal with an excess amount of chlorine gas. The chemical reaction that occurred is shown
Na + Cl2-NaCl
If the percentage yield of the reaction is 86%, what is the actual yield? Show your work, including the use of stoichiometric calculations and conversion factors

Answer 1

Answer: 125.646 g

Explanation:

$Given chemical reaction : - \\ \mathrm{Na}+\mathrm{Cl}_{2} \rightarrow \mathrm{NaCl} \\Balanced chemical equation\\ 2 \mathrm{Na}+\mathrm{Cl}_{2} \rightarrow 2 \mathrm{NaCl}$

$2 \times 23 \mathrm{~g} \text { of } \mathrm{Na} \text { produces } &=2 \times 58.44 \mathrm{~g} \text { of } \mathrm{NaCl}$

$\Rightarrow 1 \mathrm{~g} \text { of } \mathrm{Na} \text { will produce }=& \frac{2 \times 58.44 \mathrm{~g} \text { of } \mathrm{NaCl}}{2 \times 23}$

$\text { So } 57.50 \mathrm{~g} \text { of } \mathrm{Na} \text { will produce } &=\frac{2 \times 58.44}{2 \times 23} \times 57.50 \mathrm{~g} \text { of } \mathrm{NaCl} \\& \bold{\Rightarrow 146.1 \mathrm{~g}}$

$\% \text { yield }=\frac{\text { Actual yield }}{\text { Theoretical yield }} \times 100$

$\begin{aligned}\text { So actual yield } &=\frac{\% \text { yield } \times \text { Theoretical yield }}{100} \\&=\frac{86 \times 146.1}{100} \\& \bf{\Rightarrow 125.646 \mathrm{~g}}\end{aligned}$

Therefore, the actual yield for this reaction is 125.646 g