17. What volume, in liters, of hydrogen gas measured at STP can be produced from 50.0 mL of 6.0 M hydrobromic acid and an excess of zinc metal, according to the following reaction? 2HB1(aq) + Zn(s) → ZnBr2(aq) + Hz(g)
17. What volume, in liters, of hydrogen gas measured at STP can be produced from 50.0 mL of 6.0 M hydrobromic acid and an excess of zinc metal, according to the following reaction? 2HB1(aq) + Zn(s) → ZnBr2(aq) + Hz(g)
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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What volume, in liters, of hydrogen gas measured at STP can be produced from 50.0 mL of 6.0 M hydrobromic acid and an excess of zinc metal, according to the following reaction?
![**Problem 17**
What volume, in liters, of hydrogen gas measured at STP can be produced from 50.0 mL of 6.0 M hydrobromic acid and an excess of zinc metal, according to the following reaction?
\[ \text{2HBr(aq) + Zn(s) } \rightarrow \text{ ZnBr}_2\text{(aq) + H}_2\text{(g)} \]
**Explanation**
This is a stoichiometry problem involving the reaction of hydrobromic acid (HBr) with zinc (Zn) to produce zinc bromide (ZnBr₂) and hydrogen gas (H₂). To solve this, follow these steps:
1. **Calculate moles of HBr:**
- Volume of HBr = 50.0 mL = 0.050 L
- Concentration of HBr = 6.0 M
- Moles of HBr = Volume × Concentration = 0.050 L × 6.0 mol/L = 0.30 mol
2. **Use stoichiometry to find moles of H₂:**
- From the balanced equation: 2 moles of HBr produce 1 mole of H₂.
- Moles of H₂ = (0.30 mol HBr) × (1 mol H₂ / 2 mol HBr) = 0.15 mol H₂
3. **Calculate volume of H₂ at STP:**
- At STP, 1 mole of gas occupies 22.4 L
- Volume of H₂ = 0.15 mol × 22.4 L/mol = 3.36 L
Thus, 3.36 liters of hydrogen gas can be produced under these conditions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1d66f0ae-8c6e-4ff5-9763-c8b27ba74e97%2Fdfc9bd83-d0a6-4879-bb97-ab015c7a51df%2Fnw1dxhc_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 17**
What volume, in liters, of hydrogen gas measured at STP can be produced from 50.0 mL of 6.0 M hydrobromic acid and an excess of zinc metal, according to the following reaction?
\[ \text{2HBr(aq) + Zn(s) } \rightarrow \text{ ZnBr}_2\text{(aq) + H}_2\text{(g)} \]
**Explanation**
This is a stoichiometry problem involving the reaction of hydrobromic acid (HBr) with zinc (Zn) to produce zinc bromide (ZnBr₂) and hydrogen gas (H₂). To solve this, follow these steps:
1. **Calculate moles of HBr:**
- Volume of HBr = 50.0 mL = 0.050 L
- Concentration of HBr = 6.0 M
- Moles of HBr = Volume × Concentration = 0.050 L × 6.0 mol/L = 0.30 mol
2. **Use stoichiometry to find moles of H₂:**
- From the balanced equation: 2 moles of HBr produce 1 mole of H₂.
- Moles of H₂ = (0.30 mol HBr) × (1 mol H₂ / 2 mol HBr) = 0.15 mol H₂
3. **Calculate volume of H₂ at STP:**
- At STP, 1 mole of gas occupies 22.4 L
- Volume of H₂ = 0.15 mol × 22.4 L/mol = 3.36 L
Thus, 3.36 liters of hydrogen gas can be produced under these conditions.
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