To solve the problem, we need to determine the minimum volume of a \( \frac{5}{51} \, M \) aluminum sulfate solution required to produce at least 1 gram of each salt in the reaction:
\[
\text{Al}_2(\text{SO}_4)_3 + 2 \text{Ca(NO}_3)_2 \rightarrow 2 \text{Al(NO}_3)_3 + 3 \text{CaSO}_4
\]
### Step 1: Calculate the molar masses of the compounds involved.
- Molar mass of \( \text{CaSO}_4 \) = 136 g/mol
- Molar mass of \( \text{Al}_2(\text{SO}_4)_3 \) = 342 g/mol
- Molar mass of \( \text{Ca(NO}_3)_2 \) = 164 g/mol
- Molar mass of \( \text{Al(NO}_3)_3 \) = 212.996 g/mol
### Step 2: Calculate the number of moles required for 1 gram of each salt.
- For \( \text{CaSO}_4 \):
\[
\text{Number of moles} = \frac{1 \, \text{g}}{136 \, \text{g/mol}} = 0.00735 \, \text{mol}
\]
- For \( \text{Al(NO}_3)_3 \):
\[
\text{Number of moles} = \frac{1 \, \text{g}}{212.996 \, \text{g/mol}} = 0.00469 \, \text{mol}
\]
### Step 3: Relate the moles of aluminum sulfate to the moles of the products.
From the balanced equation, we see that:
- 1 mole of \( \text{Al}_2(\text{SO}_4)_3 \) produces 2 moles of \( \text{Al(NO}_3)_3 \) and 3 moles of \( \text{CaSO}_4 \).
Thus, the moles of \( \text{Al}_2(\text{SO}_4)_3 \) needed to produce \( 0.00469 \, \text{mol} \) of \( \text{Al(NO}_3)_3 \) is:
\[
\text{Moles of } \text{Al}_2(\text{SO}_4)_3 = \frac{0.00469}{2} = 0.002345 \, \text{mol}
\]
### Step 4: Calculate the mass of aluminum sulfate required.
Now, we can find the mass of \( \text{Al}_2(\text{SO}_4)_3 \):
\[
\text{Mass} = \text{Moles} \times \text{Molar mass} = 0.002345 \, \text{mol} \times 342 \, \text{g/mol} = 0.80169 \, \text{g}
\]
### Step 5: Calculate the volume of aluminum sulfate solution needed.
Using the molarity equation:
\[
\text{Molarity} = \frac{\text{Number of moles}}{\text{Volume in L}}
\]
Rearranging gives:
\[
\text{Volume in L} = \frac{\text{Number of moles}}{\text{Molarity}} = \frac{0.002345 \, \text{mol}}{\frac{5}{51} \, M}
\]
Calculating the volume:
\[
\text{Volume in L} = \frac{0.002345}{\frac{5}{51}} = 0.002345 \times \frac{51}{5} = 0.0239 \, \text{L} = 23.9 \, \text{mL}
\]
### Final Answer:
The minimum volume of \( \frac{5}{51} \, M \) aluminum sulfate solution required is approximately **24 mL**.
---
