Given the data in the table, how many grams of $\mathrm{NaCl}$ would you expect to be formed in the reaction of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ with excess $\mathrm{HCl}$ ? The molar mass of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ is $105.99 \mathrm{~g} / \mathrm{mol}$, and the molar mass of $\mathrm{NaCl}$ is $58.44 \mathrm{~g} / \mathrm{mol}$.
\begin{tabular}{|c|c|}
\hline mass of empty beaker $(\mathrm{g})$ & 84.000 \
\hline mass of beaker plus $\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{~g})$ & 85.936 \
\hline mass of $\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{~g})$ & 1.936 \
\hline
\end{tabular}
mass of $\mathrm{NaCl}$ : $\square$
$\mathrm{g}$
If $0.973 \mathrm{~g}$ of $\mathrm{NaCl}$ was produced in the experiment, calculate the percent yield of $\mathrm{NaCl}$.
percent yield: $\square$ $\%$

Question

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Identify the reaction**: The reaction between sodium carbonate ($\mathrm{Na}_{2} \mathrm{CO}_{3}$) and hydrochloric acid ($\mathrm{HCl}$) can be represented by the balanced chemical equation: $$\mathrm{Na}_{2} \mathrm{CO}_{3} + 2 \mathrm{HCl} ightarrow 2 \mathrm{NaCl} + \mathrm{H}_{2} \mathrm{O} + \mathrm{CO}_{2}$$ ‖ ‖ 2 ⋮ **Calculate moles of $\mathrm{Na}_{2} \mathrm{CO}_{3}$**: Using the given mass of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ and its molar mass: $$ ext{Mass of } \mathrm{Na}_{2} \mathrm{CO}_{3} = 1.936 \, \mathrm{g}$$ $$ ext{Molar mass of } \mathrm{Na}_{2} \mathrm{CO}_{3} = 105.99 \, \mathrm{g/mol}$$ $$ ext{Moles of } \mathrm{Na}_{2} \mathrm{CO}_{3} = \frac{1.936 \, \mathrm{g}}{105.99 \, \mathrm{g/mol}} = 0.01826 \, \mathrm{mol}$$ ‖ ‖ 3 ⋮ **Determine moles of $\mathrm{NaCl}$ produced**: According to the balanced equation, 1 mole of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ produces 2 moles of $\mathrm{NaCl}$: $$ ext{Moles of } \mathrm{NaCl} = 2 imes 0.01826 \, \mathrm{mol} = 0.03652 \, \mathrm{mol}$$ ‖ ‖ 4 ⋮ **Calculate the mass of $\mathrm{NaCl}$**: Using the molar mass of $\mathrm{NaCl}$: $$ ext{Molar mass of } \mathrm{NaCl} = 58.44 \, \mathrm{g/mol}$$ $$ ext{Mass of } \mathrm{NaCl} = 0.03652 \, \mathrm{mol} imes 58.44 \, \mathrm{g/mol} = 2.134 \, \mathrm{g}$$ ‖ ‖ 5 ⋮ **Calculate the percent yield**: Using the actual yield and theoretical yield: $$ ext{Actual yield} = 0.973 \, \mathrm{g}$$ $$ ext{Theoretical yield} = 2.134 \, \mathrm{g}$$ $$ ext{Percent yield} = \left( \frac{0.973 \, \mathrm{g}}{2.134 \, \mathrm{g}} ight) imes 100\% = 45.6\%$$ ‖ ∻Answer∻ ⚹ The mass of $\mathrm{NaCl}$ expected to be formed is $2.134 \, \mathrm{g}$, and the percent yield of $\mathrm{NaCl}$ is $45.6\%$. ⚹ ∻Key Concept∻ ⚹ Stoichiometry and percent yield calculation ⚹ ∻Explanation∻ ⚹ The problem involves using stoichiometry to determine the theoretical yield of $\mathrm{NaCl}$ from the reaction of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ with excess $\mathrm{HCl}$, and then calculating the percent yield based on the actual yield obtained from the experiment. ⚹◊What is the balanced chemical equation for the reaction of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ with excess $\mathrm{HCl}$⍭ Generate me a similar question◊

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