Question

1. if corey produces 7.3 grams of sodium carbonate in a reaction that results in a 74.0% yield, how many grams of sodium phosphate should he obtain?

Explanation:

To solve this problem, we need to know the balanced chemical equation for the reaction involving sodium carbonate and sodium phosphate. Assuming the reaction is between calcium phosphate and sodium carbonate (a common reaction to produce sodium phosphate and calcium carbonate), the balanced equation is:

$$\text{Ca}_3(\text{PO}_4)_2 + 3\text{Na}_2\text{CO}_3 ightarrow 2\text{Na}_3\text{PO}_4 + 3\text{CaCO}_3$$

Step 1: Calculate the theoretical yield of sodium carbonate

The actual yield of sodium carbonate is 7.3 g, and the percent yield is 74.0%. The formula for percent yield is:

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

Rearranging to solve for theoretical yield:

$$\text{Theoretical Yield of } \text{Na}_2\text{CO}_3 = \frac{\text{Actual Yield}}{\text{Percent Yield}} \times 100\%$$

Substituting the given values:

$$\text{Theoretical Yield of } \text{Na}_2\text{CO}_3 = \frac{7.3\ \text{g}}{74.0\%} \times 100\% = \frac{7.3\ \text{g}}{0.74} \approx 9.865\ \text{g}$$

Step 2: Calculate the moles of sodium carbonate

The molar mass of $\text{Na}_2\text{CO}_3$ is:

$$2 \times 22.99\ \text{g/mol (Na)} + 12.01\ \text{g/mol (C)} + 3 \times 16.00\ \text{g/mol (O)} = 105.99\ \text{g/mol}$$

Moles of $\text{Na}_2\text{CO}_3$:

$$\text{Moles of } \text{Na}_2\text{CO}_3 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{9.865\ \text{g}}{105.99\ \text{g/mol}} \approx 0.0931\ \text{mol}$$

Step 3: Determine the moles of sodium phosphate from the stoichiometry

From the balanced equation, 3 moles of $\text{Na}_2\text{CO}_3$ produce 2 moles of $\text{Na}_3\text{PO}_4$. So the mole ratio of $\text{Na}_3\text{PO}_4$ to $\text{Na}_2\text{CO}_3$ is $\frac{2}{3}$.

Moles of $\text{Na}_3\text{PO}_4$:

$$\text{Moles of } \text{Na}_3\text{PO}_4 = 0.0931\ \text{mol (Na}_2\text{CO}_3) \times \frac{2}{3} \approx 0.0621\ \text{mol}$$

Step 4: Calculate the mass of sodium phosphate

The molar mass of $\text{Na}_3\text{PO}_4$ is:

$$3 \times 22.99\ \text{g/mol (Na)} + 30.97\ \text{g/mol (P)} + 4 \times 16.00\ \text{g/mol (O)} = 163.94\ \text{g/mol}$$

Mass of $\text{Na}_3\text{PO}_4$:

$$\text{Mass of } \text{Na}_3\text{PO}_4 = \text{Moles} \times \text{Molar Mass} = 0.0621\ \text{mol} \times 163.94\ \text{g/mol} \approx 10.18\ \text{g}$$

Final Answer

The mass of sodium phosphate Corey should obtain is approximately $\boxed{10.2\ \text{g}}$ (rounded to three significant figures).

Answer:

To solve this problem, we need to know the balanced chemical equation for the reaction involving sodium carbonate and sodium phosphate. Assuming the reaction is between calcium phosphate and sodium carbonate (a common reaction to produce sodium phosphate and calcium carbonate), the balanced equation is:

$$\text{Ca}_3(\text{PO}_4)_2 + 3\text{Na}_2\text{CO}_3 ightarrow 2\text{Na}_3\text{PO}_4 + 3\text{CaCO}_3$$

Step 1: Calculate the theoretical yield of sodium carbonate

The actual yield of sodium carbonate is 7.3 g, and the percent yield is 74.0%. The formula for percent yield is:

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

Rearranging to solve for theoretical yield:

$$\text{Theoretical Yield of } \text{Na}_2\text{CO}_3 = \frac{\text{Actual Yield}}{\text{Percent Yield}} \times 100\%$$

Substituting the given values:

$$\text{Theoretical Yield of } \text{Na}_2\text{CO}_3 = \frac{7.3\ \text{g}}{74.0\%} \times 100\% = \frac{7.3\ \text{g}}{0.74} \approx 9.865\ \text{g}$$

Step 2: Calculate the moles of sodium carbonate

The molar mass of $\text{Na}_2\text{CO}_3$ is:

$$2 \times 22.99\ \text{g/mol (Na)} + 12.01\ \text{g/mol (C)} + 3 \times 16.00\ \text{g/mol (O)} = 105.99\ \text{g/mol}$$

Moles of $\text{Na}_2\text{CO}_3$:

$$\text{Moles of } \text{Na}_2\text{CO}_3 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{9.865\ \text{g}}{105.99\ \text{g/mol}} \approx 0.0931\ \text{mol}$$

Step 3: Determine the moles of sodium phosphate from the stoichiometry

From the balanced equation, 3 moles of $\text{Na}_2\text{CO}_3$ produce 2 moles of $\text{Na}_3\text{PO}_4$. So the mole ratio of $\text{Na}_3\text{PO}_4$ to $\text{Na}_2\text{CO}_3$ is $\frac{2}{3}$.

Moles of $\text{Na}_3\text{PO}_4$:

$$\text{Moles of } \text{Na}_3\text{PO}_4 = 0.0931\ \text{mol (Na}_2\text{CO}_3) \times \frac{2}{3} \approx 0.0621\ \text{mol}$$

Step 4: Calculate the mass of sodium phosphate

The molar mass of $\text{Na}_3\text{PO}_4$ is:

$$3 \times 22.99\ \text{g/mol (Na)} + 30.97\ \text{g/mol (P)} + 4 \times 16.00\ \text{g/mol (O)} = 163.94\ \text{g/mol}$$

Mass of $\text{Na}_3\text{PO}_4$:

$$\text{Mass of } \text{Na}_3\text{PO}_4 = \text{Moles} \times \text{Molar Mass} = 0.0621\ \text{mol} \times 163.94\ \text{g/mol} \approx 10.18\ \text{g}$$

Final Answer

The mass of sodium phosphate Corey should obtain is approximately $\boxed{10.2\ \text{g}}$ (rounded to three significant figures).