What volume in liters of H, gas would be produced by the complete reaction of 2.93 g of Al solid at STP according to the following reaction? Remember 1 mol of an ideal gas has a volume of 22.4 L at STP. 2 Al (s) + 6 HCI (aq) → 2 AICI, (aq) + 3 H₂(g)
What volume in liters of H, gas would be produced by the complete reaction of 2.93 g of Al solid at STP according to the following reaction? Remember 1 mol of an ideal gas has a volume of 22.4 L at STP. 2 Al (s) + 6 HCI (aq) → 2 AICI, (aq) + 3 H₂(g)
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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![### Calculating Volume of Hydrogen Gas Produced in a Reaction
In this educational section, we will determine the volume of hydrogen gas (\(H_2\)) produced from the reaction of aluminum (\(Al\)) with hydrochloric acid (\(HCl\)) at standard temperature and pressure (STP).
#### Problem Statement
What volume in liters of \(H_2\) gas would be produced by the complete reaction of **2.93 g of Al** solid at STP according to the following reaction? Remember that **1 mol** of an ideal gas has a volume of **22.4 L** at STP.
\[2 Al (s) + 6 HCl (aq) \rightarrow 2 AlCl_3 (aq) + 3 H_2 (g)\]
#### Setting Up the Calculation
To solve this problem, we will use the stoichiometry of the balanced chemical equation.
1. **Calculate moles of \(Al\)**:
Given:
- Mass of \(Al\) = 2.93 g
- Molar mass of \(Al\) = 26.98 g/mol
\[\text{Moles of } Al = \frac{2.93 \, \text{g}}{26.98 \, \text{g/mol}} = 0.1087 \, \text{mol}\]
2. **Use the stoichiometric ratio** between \(Al\) and \(H_2\) from the balanced equation:
- From the equation, 2 moles of \(Al\) produce 3 moles of \(H_2\).
\[\text{Moles of } H_2 = 0.1087 \, \text{mol Al} \times \frac{3 \, \text{mol H}_2}{2 \, \text{mol Al}} = 0.1630 \, \text{mol H}_2\]
3. **Calculate the volume of \(H_2\) gas** at STP:
- Using the ideal gas law where 1 mole of an ideal gas at STP occupies 22.4 liters,
\[\text{Volume of } H_2 = 0.1630 \, \text{mol} \times 22.4 \, \text{L/mol} = 3.65 \, \text](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffbb15d99-77db-4308-835f-20d2902f98e0%2F78f4fac3-4a85-4e99-abe4-7183f4da86e1%2Fna225ja_processed.png&w=3840&q=75)
Transcribed Image Text:### Calculating Volume of Hydrogen Gas Produced in a Reaction
In this educational section, we will determine the volume of hydrogen gas (\(H_2\)) produced from the reaction of aluminum (\(Al\)) with hydrochloric acid (\(HCl\)) at standard temperature and pressure (STP).
#### Problem Statement
What volume in liters of \(H_2\) gas would be produced by the complete reaction of **2.93 g of Al** solid at STP according to the following reaction? Remember that **1 mol** of an ideal gas has a volume of **22.4 L** at STP.
\[2 Al (s) + 6 HCl (aq) \rightarrow 2 AlCl_3 (aq) + 3 H_2 (g)\]
#### Setting Up the Calculation
To solve this problem, we will use the stoichiometry of the balanced chemical equation.
1. **Calculate moles of \(Al\)**:
Given:
- Mass of \(Al\) = 2.93 g
- Molar mass of \(Al\) = 26.98 g/mol
\[\text{Moles of } Al = \frac{2.93 \, \text{g}}{26.98 \, \text{g/mol}} = 0.1087 \, \text{mol}\]
2. **Use the stoichiometric ratio** between \(Al\) and \(H_2\) from the balanced equation:
- From the equation, 2 moles of \(Al\) produce 3 moles of \(H_2\).
\[\text{Moles of } H_2 = 0.1087 \, \text{mol Al} \times \frac{3 \, \text{mol H}_2}{2 \, \text{mol Al}} = 0.1630 \, \text{mol H}_2\]
3. **Calculate the volume of \(H_2\) gas** at STP:
- Using the ideal gas law where 1 mole of an ideal gas at STP occupies 22.4 liters,
\[\text{Volume of } H_2 = 0.1630 \, \text{mol} \times 22.4 \, \text{L/mol} = 3.65 \, \text
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