How many liters of hydrogen gas would be produced by the complete reaction of 2.93 g of Al at STP according to the following reaction? Remember 1 mol of an ideal gas has a volume of 22.4 Lat STP 2 Al (s) + 6 HCI (aq) → 2 AICI, (aq) + 3 H2 (9) STARTING AMOUNT ADD FACTOR ANSWER RESET *( ) 6.022 x 10 22.4 2.02 1.62 2.93 1 2 0.329 7.30 26.98 3.65 6 mol H2 mol Al g/mol Al g H2 g/mol H2 LH2 LAI g Al 3.
How many liters of hydrogen gas would be produced by the complete reaction of 2.93 g of Al at STP according to the following reaction? Remember 1 mol of an ideal gas has a volume of 22.4 Lat STP 2 Al (s) + 6 HCI (aq) → 2 AICI, (aq) + 3 H2 (9) STARTING AMOUNT ADD FACTOR ANSWER RESET *( ) 6.022 x 10 22.4 2.02 1.62 2.93 1 2 0.329 7.30 26.98 3.65 6 mol H2 mol Al g/mol Al g H2 g/mol H2 LH2 LAI g Al 3.
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![**Calculating the Volume of Hydrogen Gas Produced in a Chemical Reaction**
**Question:**
How many liters of hydrogen gas would be produced by the complete reaction of 2.93 g of Al at STP according to the following reaction? Remember 1 mol of an ideal gas has a volume of 22.4 L at STP.
\[ \text{2 Al (s) + 6 HCl (aq) } \rightarrow \text{ 2 AlCl}_3 (aq) + \text{3 H}_2 \text{ (g)} \]
**Instructions:**
1. **Starting Amount & Reaction Setup:**
- You are given 2.93 grams of aluminum (Al).
- The reaction mentioned is:
\[ \text{2 Al (s) + 6 HCl (aq) } \rightarrow \text{ 2 AlCl}_3 (aq) + \text{3 H}_2 \text{ (g)} \]
2. **Key Values to Use:**
- Molar mass of Al: ≈ 26.98 g/mol.
- Ideal gas volume at STP: 22.4 L/mol.
3. **Calculation Steps:**
1. **Convert grams of Al to moles:**
\[ \text{Moles of Al} = \frac{\text{grams of Al}}{\text{molar mass of Al}} = \frac{2.93 \text{ g}}{26.98 \text{ g/mol}} \]
2. **Use the stoichiometric relationship from the balanced equation to find moles of \(\text{H}_2\):**
\[ 2 \text{ mol Al} \rightarrow 3 \text{ mol H}_2 \]
\[ \text{Moles of } \text{H}_2 = \left( \text{moles of Al} \right) \times \frac{3 \text{ mol H}_2}{2 \text{ mol Al}} \]
3. **Convert moles of \(\text{H}_2\) to liters of \(\text{H}_2\) at STP:**
\[ \text{Volume of } \text{H}_2 = \left( \text{moles of } \text{H}_2 \right) \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F44982666-d339-4885-b9e9-36706f6e8dd1%2Fe65f0a1b-9174-4d68-9df9-5d6f5ce35e36%2F2577p4_processed.png&w=3840&q=75)
Transcribed Image Text:**Calculating the Volume of Hydrogen Gas Produced in a Chemical Reaction**
**Question:**
How many liters of hydrogen gas would be produced by the complete reaction of 2.93 g of Al at STP according to the following reaction? Remember 1 mol of an ideal gas has a volume of 22.4 L at STP.
\[ \text{2 Al (s) + 6 HCl (aq) } \rightarrow \text{ 2 AlCl}_3 (aq) + \text{3 H}_2 \text{ (g)} \]
**Instructions:**
1. **Starting Amount & Reaction Setup:**
- You are given 2.93 grams of aluminum (Al).
- The reaction mentioned is:
\[ \text{2 Al (s) + 6 HCl (aq) } \rightarrow \text{ 2 AlCl}_3 (aq) + \text{3 H}_2 \text{ (g)} \]
2. **Key Values to Use:**
- Molar mass of Al: ≈ 26.98 g/mol.
- Ideal gas volume at STP: 22.4 L/mol.
3. **Calculation Steps:**
1. **Convert grams of Al to moles:**
\[ \text{Moles of Al} = \frac{\text{grams of Al}}{\text{molar mass of Al}} = \frac{2.93 \text{ g}}{26.98 \text{ g/mol}} \]
2. **Use the stoichiometric relationship from the balanced equation to find moles of \(\text{H}_2\):**
\[ 2 \text{ mol Al} \rightarrow 3 \text{ mol H}_2 \]
\[ \text{Moles of } \text{H}_2 = \left( \text{moles of Al} \right) \times \frac{3 \text{ mol H}_2}{2 \text{ mol Al}} \]
3. **Convert moles of \(\text{H}_2\) to liters of \(\text{H}_2\) at STP:**
\[ \text{Volume of } \text{H}_2 = \left( \text{moles of } \text{H}_2 \right) \
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