Determine the mass in grams of HCI that can react with 0.750 g of Al(OH), according to the following reaction Al(OH),(s) + 3 HCl(aq) → AICI, (aq) + 3 H,O(aq)
Determine the mass in grams of HCI that can react with 0.750 g of Al(OH), according to the following reaction Al(OH),(s) + 3 HCl(aq) → AICI, (aq) + 3 H,O(aq)
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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![**Chemical Reaction Calculation**
**Objective:**
Determine the mass in grams of HCl that can react with 0.750 g of Al(OH)₃ according to the following reaction:
\[
\text{Al(OH)}_3(s) + 3 \text{HCl}(aq) \rightarrow \text{AlCl}_3(aq) + 3 \text{H}_2\text{O}(aq)
\]
**Steps for Calculation:**
1. **Starting Amount** - The initial mass of Al(OH)₃ given is 0.750 g.
2. **Molar Mass Conversion** - Use the molar mass of Al(OH)₃ to convert grams to moles:
- Molar mass of Al(OH)₃ = 78.00 g/mol
- Thus, moles of Al(OH)₃ = \( \frac{0.750 \text{ g}}{78.00 \text{ g/mol}} \)
3. **Stoichiometric Calculation** - Use stoichiometry to convert moles of Al(OH)₃ to moles of HCl:
- The balanced equation indicates that 1 mole of Al(OH)₃ reacts with 3 moles of HCl.
- Thus, moles of HCl required = 3 × moles of Al(OH)₃
4. **Molar Mass Conversion** - Convert moles of HCl to grams:
- Molar mass of HCl = 36.46 g/mol
- Thus, grams of HCl = moles of HCl × 36.46 g/mol
**Summary of Calculations:**
\[
\text{Starting amount:} \quad 0.750 \text{ g Al(OH)}_3
\]
\[
\text{Conversion:} \quad \left( \frac{1 \text{ mol Al(OH)}_3}{78.00 \text{ g Al(OH)}_3} \right) = \text{ 0.00962 mol Al(OH)}_3
\]
\[
\text{Stoichiometry:} \quad (0.00962 \text{ mol Al(OH)}_3) \times 3 (\text{HCl Al(OH)}_3 \text{ ratio from equation}) = 0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F20214f54-b261-4746-a2af-8e52bbd8e109%2F5a8914dd-2a6f-4562-a998-21a2fcef0f5b%2Fn49uzti_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Chemical Reaction Calculation**
**Objective:**
Determine the mass in grams of HCl that can react with 0.750 g of Al(OH)₃ according to the following reaction:
\[
\text{Al(OH)}_3(s) + 3 \text{HCl}(aq) \rightarrow \text{AlCl}_3(aq) + 3 \text{H}_2\text{O}(aq)
\]
**Steps for Calculation:**
1. **Starting Amount** - The initial mass of Al(OH)₃ given is 0.750 g.
2. **Molar Mass Conversion** - Use the molar mass of Al(OH)₃ to convert grams to moles:
- Molar mass of Al(OH)₃ = 78.00 g/mol
- Thus, moles of Al(OH)₃ = \( \frac{0.750 \text{ g}}{78.00 \text{ g/mol}} \)
3. **Stoichiometric Calculation** - Use stoichiometry to convert moles of Al(OH)₃ to moles of HCl:
- The balanced equation indicates that 1 mole of Al(OH)₃ reacts with 3 moles of HCl.
- Thus, moles of HCl required = 3 × moles of Al(OH)₃
4. **Molar Mass Conversion** - Convert moles of HCl to grams:
- Molar mass of HCl = 36.46 g/mol
- Thus, grams of HCl = moles of HCl × 36.46 g/mol
**Summary of Calculations:**
\[
\text{Starting amount:} \quad 0.750 \text{ g Al(OH)}_3
\]
\[
\text{Conversion:} \quad \left( \frac{1 \text{ mol Al(OH)}_3}{78.00 \text{ g Al(OH)}_3} \right) = \text{ 0.00962 mol Al(OH)}_3
\]
\[
\text{Stoichiometry:} \quad (0.00962 \text{ mol Al(OH)}_3) \times 3 (\text{HCl Al(OH)}_3 \text{ ratio from equation}) = 0
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