Questions: Name: Lab Partner: Experiment 7 Post-Lab Sheet Results Data (4 pts) - mass of crucible and cover: 27.5651 g - mass of crucible, cover and NaHCO3: 51.5510 g - mass of crucible, cover and product: 44.4523 g - theoretical yield of Na2CO3: - theoretical yield of Na2O: - mass of product: - identity of product: - % yield of product: (2 pts) Show how you calculated the theoretical yield of Na2CO3. Include units! (1 pt) Show how you calculated the % yield:

Name: Lab Partner: Experiment 7 Post-Lab Sheet Results Data (4 pts) - mass of crucible and cover: 27.5651 g - mass of crucible, cover and NaHCO3: 51.5510 g - mass of crucible, cover and product: 44.4523 g - theoretical yield of Na2CO3: - theoretical yield of Na2O: - mass of product: - identity of product: - % yield of product: (2 pts) Show how you calculated the theoretical yield of Na2CO3. Include units! (1 pt) Show how you calculated the % yield:
Transcript text: Name: Lab Partner: Experiment 7 Post-Lab Sheet Results Data (4 pts) \begin{tabular}{|l|l|} \hline mass of crucible and cover & 27.5651 g \\ \hline mass of crucible, cover and $\mathrm{NaHCO}_{3}$ & 51.5510 g \\ \hline mass of crucible, cover and product & 44.4523 g \\ \hline theoretical yield of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ & \\ \hline theoretical yield of $\mathrm{Na}_{2} \mathrm{O}$ & \\ \hline mass of product & \\ \hline identity of product & \\ \hline \% yield of product & \\ \hline \end{tabular} (2 pts) Show how you calculated the theoretical yield of $\mathrm{Na}_{2} \mathrm{CO}_{3}$. Include units! (1 pt) Show how you calculated the \% yield:
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Solution

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Solution Steps

Step 1: Calculate the mass of $\mathrm{NaHCO}_{3}$ used

To find the mass of $\mathrm{NaHCO}_{3}$ used, subtract the mass of the crucible and cover from the mass of the crucible, cover, and $\mathrm{NaHCO}_{3}$.

\[ \text{mass of } \mathrm{NaHCO}_{3} = 51.5510 \, \text{g} - 27.5651 \, \text{g} = 23.9859 \, \text{g} \]

Step 2: Calculate the moles of $\mathrm{NaHCO}_{3}$ used

Using the molar mass of $\mathrm{NaHCO}_{3}$ (84.0066 g/mol), calculate the moles of $\mathrm{NaHCO}_{3}$.

\[ \text{moles of } \mathrm{NaHCO}_{3} = \frac{23.9859 \, \text{g}}{84.0066 \, \text{g/mol}} = 0.2855 \, \text{mol} \]

Step 3: Calculate the theoretical yield of $\mathrm{Na}_{2}\mathrm{CO}_{3}$

The balanced chemical equation for the decomposition of $\mathrm{NaHCO}_{3}$ is:

\[ 2 \mathrm{NaHCO}_{3} \rightarrow \mathrm{Na}_{2}\mathrm{CO}_{3} + \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O} \]

From the equation, 2 moles of $\mathrm{NaHCO}_{3}$ produce 1 mole of $\mathrm{Na}_{2}\mathrm{CO}_{3}$. Therefore, the moles of $\mathrm{Na}_{2}\mathrm{CO}_{3}$ produced are:

\[ \text{moles of } \mathrm{Na}_{2}\mathrm{CO}_{3} = \frac{0.2855 \, \text{mol}}{2} = 0.1428 \, \text{mol} \]

Using the molar mass of $\mathrm{Na}_{2}\mathrm{CO}_{3}$ (105.9888 g/mol), calculate the theoretical yield:

\[ \text{theoretical yield of } \mathrm{Na}_{2}\mathrm{CO}_{3} = 0.1428 \, \text{mol} \times 105.9888 \, \text{g/mol} = 15.1347 \, \text{g} \]

\[ \boxed{\text{Theoretical yield of } \mathrm{Na}_{2}\mathrm{CO}_{3} = 15.1347 \, \text{g}} \]

Step 4: Calculate the mass of the product

To find the mass of the product, subtract the mass of the crucible and cover from the mass of the crucible, cover, and product.

\[ \text{mass of product} = 44.4523 \, \text{g} - 27.5651 \, \text{g} = 16.8872 \, \text{g} \]

\[ \boxed{\text{Mass of product} = 16.8872 \, \text{g}} \]

Step 5: Calculate the \% yield

The percent yield is calculated using the formula:

\[ \% \text{ yield} = \left( \frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100\% \]

Using the actual yield (mass of product) and the theoretical yield of $\mathrm{Na}_{2}\mathrm{CO}_{3}$:

\[ \% \text{ yield} = \left( \frac{16.8872 \, \text{g}}{15.1347 \, \text{g}} \right) \times 100\% = 111.58\% \]

\[ \boxed{\% \text{ yield} = 111.58\%} \]

Final Answer

\[ \boxed{\text{Theoretical yield of } \mathrm{Na}_{2}\mathrm{CO}_{3} = 15.1347 \, \text{g}} \] \[ \boxed{\text{Mass of product} = 16.8872 \, \text{g}} \] \[ \boxed{\% \text{ yield} = 111.58\%} \]

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