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)
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)
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3faa4e08-2098-47c7-aa8f-039b05387853%2F98a9ba99-4b57-48b4-abab-a4e1034cab2b%2Fry2226p.jpeg&w=3840&q=75)
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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