What volume (in mL) of 0.300 M HCI would be required to completely react with 5.55 g of Al in the following chemical reaction? 2 Al(s) + 6 HCI(aq) → 2 AICIls (aq) + 3 H2 (g)

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**Title: Calculating Required Volume of HCl for Complete Reaction with Aluminum**

**Question 23 of 25:**

**Problem Statement:**
What volume (in mL) of 0.300 M HCl would be required to completely react with 5.55 g of Al in the following chemical reaction?

\[ 2 \text{Al(s)} + 6 \text{HCl(aq)} \rightarrow 2 \text{AlCl}_3 (\text{aq}) + 3 \text{H}_2 (\text{g}) \]

**Calculation Steps:**

1. **Write the Balanced Equation:**
   \[ 2 \text{Al(s)} + 6 \text{HCl(aq)} \rightarrow 2 \text{AlCl}_3 (\text{aq}) + 3 \text{H}_2 (\text{g}) \]

2. **Determine Moles of Al:**
   - Molar mass of \(\text{Al} = 26.98 \, \text{g/mol}\)
   - Moles of \(\text{Al} = \frac{5.55 \, \text{g}}{26.98 \, \text{g/mol}}\)

3. **Use Stoichiometry to Find Moles of HCl Needed:**
   - From the balanced equation, \(2 \text{ moles Al}\) reacts with \(6 \text{ moles HCl}\).
   - Calculate the moles of \(\text{HCl}\) needed for complete reaction.

4. **Calculate the Volume of HCl Solution:**
   - Molarity (\(M\)) of HCl solution = 0.300 M.
   - Use the formula: \[ \text{Volume (L)} = \frac{\text{Moles of HCl}}{\text{Molarity of HCl}} \]
   - Convert volume from liters to milliliters.

**Answer Input:**
A blank space is provided next to "mL" for the calculated volume of HCl.

**Interactive Component:**
A calculator interface is displayed on the right side of the screen, allowing users to enter numbers and perform calculations as needed.

**Conclusion:**
By following the steps and performing the calculations, students can determine the volume of 0.300 M HCl required to completely react with the given amount of aluminum.
Transcribed Image Text:**Title: Calculating Required Volume of HCl for Complete Reaction with Aluminum** **Question 23 of 25:** **Problem Statement:** What volume (in mL) of 0.300 M HCl would be required to completely react with 5.55 g of Al in the following chemical reaction? \[ 2 \text{Al(s)} + 6 \text{HCl(aq)} \rightarrow 2 \text{AlCl}_3 (\text{aq}) + 3 \text{H}_2 (\text{g}) \] **Calculation Steps:** 1. **Write the Balanced Equation:** \[ 2 \text{Al(s)} + 6 \text{HCl(aq)} \rightarrow 2 \text{AlCl}_3 (\text{aq}) + 3 \text{H}_2 (\text{g}) \] 2. **Determine Moles of Al:** - Molar mass of \(\text{Al} = 26.98 \, \text{g/mol}\) - Moles of \(\text{Al} = \frac{5.55 \, \text{g}}{26.98 \, \text{g/mol}}\) 3. **Use Stoichiometry to Find Moles of HCl Needed:** - From the balanced equation, \(2 \text{ moles Al}\) reacts with \(6 \text{ moles HCl}\). - Calculate the moles of \(\text{HCl}\) needed for complete reaction. 4. **Calculate the Volume of HCl Solution:** - Molarity (\(M\)) of HCl solution = 0.300 M. - Use the formula: \[ \text{Volume (L)} = \frac{\text{Moles of HCl}}{\text{Molarity of HCl}} \] - Convert volume from liters to milliliters. **Answer Input:** A blank space is provided next to "mL" for the calculated volume of HCl. **Interactive Component:** A calculator interface is displayed on the right side of the screen, allowing users to enter numbers and perform calculations as needed. **Conclusion:** By following the steps and performing the calculations, students can determine the volume of 0.300 M HCl required to completely react with the given amount of aluminum.
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