Multiple Choice
Consider the reaction 2HBr(g) → H2(g) + Br2(g). In the first 20.0 s of this reaction, the concentration of HBr dropped from 0.580 M to 0.534 M. Calculate the average rate of the reaction during this time interval in M/s.
3. Chemical Reactions
Stoichiometry
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Determine the theoretical yield of H2S (in moles) if 48 mol Al2S3 and 48 mol H2O are reacted according to the following balanced reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H2S(g).
A
72 moles
B
24 moles
C
48 moles
D
96 moles
Verified step by step guidance1
Identify the balanced chemical equation: \( \text{Al}_2\text{S}_3(s) + 6 \text{H}_2\text{O}(l) \rightarrow 2 \text{Al(OH)}_3(s) + 3 \text{H}_2\text{S}(g) \). This equation shows the stoichiometric relationships between reactants and products.
Determine the limiting reactant by comparing the mole ratio of \( \text{Al}_2\text{S}_3 \) and \( \text{H}_2\text{O} \) to the coefficients in the balanced equation. The balanced equation requires 1 mole of \( \text{Al}_2\text{S}_3 \) to react with 6 moles of \( \text{H}_2\text{O} \).
Calculate the moles of \( \text{H}_2\text{O} \) needed to completely react with 48 moles of \( \text{Al}_2\text{S}_3 \). Using the stoichiometry from the balanced equation, you need \( 48 \text{ mol Al}_2\text{S}_3 \times \frac{6 \text{ mol H}_2\text{O}}{1 \text{ mol Al}_2\text{S}_3} \).
Compare the calculated moles of \( \text{H}_2\text{O} \) needed with the available moles (48 moles). If the calculated moles exceed the available moles, \( \text{H}_2\text{O} \) is the limiting reactant; otherwise, \( \text{Al}_2\text{S}_3 \) is the limiting reactant.
Use the limiting reactant to calculate the theoretical yield of \( \text{H}_2\text{S} \). If \( \text{H}_2\text{O} \) is limiting, use \( 48 \text{ mol H}_2\text{O} \times \frac{3 \text{ mol H}_2\text{S}}{6 \text{ mol H}_2\text{O}} \). If \( \text{Al}_2\text{S}_3 \) is limiting, use \( 48 \text{ mol Al}_2\text{S}_3 \times \frac{3 \text{ mol H}_2\text{S}}{1 \text{ mol Al}_2\text{S}_3} \).
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