Question

What is the amount of Al2S3 needed to produce 200g of Al(OH)3 in the reaction Al2S3 + 6H2O --> 2Al(OH)3 + 3H2S with a 90% yield?

Answer 1

To calculate the amount of \( \text{Al}_2\text{S}_3 \) needed to produce 200 g of \( \text{Al(OH)}_3 \) with a 90% yield, we first need to find out how much \( \text{Al(OH)}_3 \) is produced theoretically based on the stoichiometry of the reaction:

\[
\text{Al}_2\text{S}_3 + 6\text{H}_2\text{O} \rightarrow 2\text{Al(OH)}_3 + 3\text{H}_2\text{S}
\]

### Step 1: Calculate the molar mass of \( \text{Al(OH)}_3 \)

- Aluminum (Al): 26.98 g/mol
- Oxygen (O): 16.00 g/mol
- Hydrogen (H): 1.01 g/mol

Calculating the molar mass of \( \text{Al(OH)}_3 \):

\[
\text{Molar mass of } \text{Al(OH)}_3 = 26.98 + 3(16.00 + 1.01) = 26.98 + 3(17.01) = 26.98 + 51.03 = 78.01 \text{ g/mol}
\]

### Step 2: Calculate the moles of \( \text{Al(OH)}_3 \) produced from 200 g

\[
\text{Moles of } \text{Al(OH)}_3 = \frac{200 \text{ g}}{78.01 \text{ g/mol}} \approx 2.564 \text{ mol}
\]

### Step 3: Calculate the theoretical moles of \( \text{Al}_2\text{S}_3 \) needed

From the balanced reaction, we see that 1 mole of \( \text{Al}_2\text{S}_3 \) produces 2 moles of \( \text{Al(OH)}_3 \). Thus, to find the moles of \( \text{Al}_2\text{S}_3 \) needed for 2.564 moles of \( \text{Al(OH)}_3 \):

\[
\text{Moles of } \text{Al}_2\text{S}_3 = \frac{2.564 \text{ mol } \text{Al(OH)}_3}{2} = 1.282 \text{ mol }
\]

### Step 4: Adjust for yield

Since the yield of the reaction is 90%, the actual amount we need to produce these moles of \( \text{Al(OH)}_3 \) will be higher. We account for yield as follows:

\[
\text{Moles of } \text{Al}_2\text{S}_3 \text{ needed} = \frac{1.282 \text{ mol}}{0.90} \approx 1.422 \text{ mol}
\]

### Step 5: Calculate the molar mass of \( \text{Al}_2\text{S}_3 \)

To find the mass of \( \text{Al}_2\text{S}_3 \), we first calculate its molar mass.

- Aluminum (Al): 26.98 g/mol (2 Al)
- Sulfur (S): 32.07 g/mol (3 S)

Calculating the molar mass of \( \text{Al}_2\text{S}_3 \):

\[
\text{Molar mass of } \text{Al}_2\text{S}_3 = 2(26.98) + 3(32.07) = 53.96 + 96.21 = 150.17 \text{ g/mol}
\]

### Step 6: Calculate the mass of \( \text{Al}_2\text{S}_3 \) required

Now we calculate the mass of \( \text{Al}_2\text{S}_3 \) needed:

\[
\text{Mass of } \text{Al}_2\text{S}_3 = \text{Moles} \times \text{Molar mass} = 1.422 \text{ mol} \times 150.17 \text{ g/mol} \approx 213.03 \text{ g}
\]

### Conclusion

The amount of \( \text{Al}_2\text{S}_3 \) needed to produce 200 g of \( \text{Al(OH)}_3 \) at a 90% yield is approximately **213.03 g**.