Determine the theoretical yield in H2S (in moles) of 16 mol Al2S3 and 16 mol H2O are reacted according to the following balanced reaction. A possibly useful molar mass is Al2S3 = 150.17g/mol.
Al2S3(s) + 6H2O(l) –> 2Al(OH)3(s) + 3H2S(g)
The correct answer and explanation is :
To determine the theoretical yield of hydrogen sulfide (H₂S) in moles when 16 moles of aluminum sulfide (Al₂S₃) and 16 moles of water (H₂O) are reacted, we first need to focus on the stoichiometric relationships given in the balanced chemical equation:
$$
\text{Al}_2\text{S}_3(s) + 6\text{H}_2\text{O}(l) \rightarrow 2\text{Al(OH)}_3(s) + 3\text{H}_2\text{S}(g)
$$
Step 1: Analyze the Stoichiometric Ratios
From the balanced equation, the mole ratio between Al₂S₃ and H₂S is as follows:
- 1 mole of Al₂S₃ produces 3 moles of H₂S.
Similarly, the mole ratio between H₂O and H₂S is:
- 6 moles of H₂O produce 3 moles of H₂S.
Step 2: Determine the Limiting Reactant
We are given 16 moles of Al₂S₃ and 16 moles of H₂O. To identify the limiting reactant, we need to compare the amount of H₂S that can be produced by each reactant.
For Al₂S₃:
The mole ratio of Al₂S₃ to H₂S is 1:3. Therefore, 16 moles of Al₂S₃ will produce:
$$
16 \, \text{mol Al}_2\text{S}_3 \times \frac{3 \, \text{mol H}_2\text{S}}{1 \, \text{mol Al}_2\text{S}_3} = 48 \, \text{mol H}_2\text{S}
$$
For H₂O:
The mole ratio of H₂O to H₂S is 6:3 (or 2:1). Therefore, 16 moles of H₂O will produce:
$$
16 \, \text{mol H}_2\text{O} \times \frac{3 \, \text{mol H}_2\text{S}}{6 \, \text{mol H}_2\text{O}} = 8 \, \text{mol H}_2\text{S}
$$
Step 3: Identify the Limiting Reactant
The amount of H₂S produced by Al₂S₃ (48 moles) is greater than the amount produced by H₂O (8 moles). Therefore, H₂O is the limiting reactant.
Step 4: Calculate the Theoretical Yield of H₂S
Since H₂O is the limiting reactant, the theoretical yield of H₂S is determined by the amount of H₂O available. From the calculations above, 16 moles of H₂O will produce 8 moles of H₂S.
Thus, the theoretical yield of H₂S is 8 moles.
Conclusion:
The theoretical yield of H₂S when 16 moles of Al₂S₃ and 16 moles of H₂O react is 8 moles. This is because H₂O is the limiting reactant, and its availability dictates the maximum amount of H₂S that can be produced.