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Determine the mass in grams of HCI that can react with 0.750 g of Al(OH), açcording to the following reaction Al(OH),(s) + 3 HCI(aq) AICI (aq)+ 3 H,O(aq) STARTING AMOUNT

Determine the mass in grams of HCI that can react with 0.750 g of Al(OH), açcording to the following reaction Al(OH),(s) + 3 HCI(aq) AICI (aq)+ 3 H,O(aq) STARTING AMOUNT

Chemistry
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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**Problem Statement:** Determine the mass in grams of HCl that can react with 0.750 g of Al(OH)₃ according to the following reaction: Al(OH)₃(s) + 3 HCl(aq) → AlCl₃(aq) + 3 H₂O(aq) **Solution Process:** 1. **Starting Amount:** - The starting material is Al(OH)₃ with a mass of 0.750 g. 2. **Chemical Reaction:** - The balanced chemical equation is provided to determine the stoichiometry of the reaction. 3. **Conversion Factors:** - Different boxes are available for entering conversion factors in a structured manner: - Top row, left to right: Typically used to convert grams to moles or moles to moles. - Bottom row, left to right: Typically used for mole to gram conversions or other necessary calculations. - Conversion factors include: - Molar masses and Avogadro's number: 6.022 × 10²³ - Relevant numerical values: 133.33, 0.0288, 1, 0.750, 36.46, 3, 0.351, 9.60 × 10⁻³, 6.022 × 10²³, 1.05, 0.117, 78.00, 18.02, 2.25 4. **Additional Calculation Tools:** - Add or modify conversion factors using the "( )" button for more complex stoichiometric conversions. 5. **Answer and Reset:** - A box for the computed answer, and a reset button to clear entries and start over. 6. **Other Options:** - Identifications like g AlCl₃, mol AlCl₃, mol HCl, g HCl, mol H₂O, g H₂O, and g Al(OH)₃ highlight what each numerical input/output might represent. This setup guides users through a stoichiometric calculation, helping determine how much HCl is required to react completely with a given amount of Al(OH)₃.
Transcribed Image Text:**Problem Statement:** Determine the mass in grams of HCl that can react with 0.750 g of Al(OH)₃ according to the following reaction: Al(OH)₃(s) + 3 HCl(aq) → AlCl₃(aq) + 3 H₂O(aq) **Solution Process:** 1. **Starting Amount:** - The starting material is Al(OH)₃ with a mass of 0.750 g. 2. **Chemical Reaction:** - The balanced chemical equation is provided to determine the stoichiometry of the reaction. 3. **Conversion Factors:** - Different boxes are available for entering conversion factors in a structured manner: - Top row, left to right: Typically used to convert grams to moles or moles to moles. - Bottom row, left to right: Typically used for mole to gram conversions or other necessary calculations. - Conversion factors include: - Molar masses and Avogadro's number: 6.022 × 10²³ - Relevant numerical values: 133.33, 0.0288, 1, 0.750, 36.46, 3, 0.351, 9.60 × 10⁻³, 6.022 × 10²³, 1.05, 0.117, 78.00, 18.02, 2.25 4. **Additional Calculation Tools:** - Add or modify conversion factors using the "( )" button for more complex stoichiometric conversions. 5. **Answer and Reset:** - A box for the computed answer, and a reset button to clear entries and start over. 6. **Other Options:** - Identifications like g AlCl₃, mol AlCl₃, mol HCl, g HCl, mol H₂O, g H₂O, and g Al(OH)₃ highlight what each numerical input/output might represent. This setup guides users through a stoichiometric calculation, helping determine how much HCl is required to react completely with a given amount of Al(OH)₃.
### Stoichiometry Problem: Reaction of Aluminum with Sulfuric Acid **Objective: Calculate Moles of \( H_2SO_4 \) Required** **Problem Statement:** What quantity of moles of \( H_2SO_4 \) is required to completely react with 7.20 moles of \( Al \), according to the balanced chemical equation: \[ 2 \, Al(s) + 3 \, H_2SO_4(aq) \rightarrow Al_2(SO_4)_3(aq) + 3 \, H_2(g) \] **Calculation Setup:** 1. **Starting Amount** - A box labeled "STARTING AMOUNT" indicates where to enter the initial quantity, in this case, 7.20 moles of \( Al \). 2. **Stoichiometric Conversion** - A visual setup for multiplication and addition factors helps to perform conversion from moles of \( Al \) to moles of \( H_2SO_4 \). \[ \begin{array}{|c|c|} \hline \text{Starting Moles} & \text{Conversion} \\ \hline \text{7.20 mol Al} & \left( \frac{3 \text{ mol } H_2SO_4}{2 \text{ mol } Al} \right) \\ \hline \end{array} \] 3. **Conversion Factors** - Various numeric buttons are available for factors such as 1, 98.08, 10.8, 26.98, and others, including Avogadro's number (6.022 × 10²³) for conversions. 4. **Result Calculation** - Complete the calculation by multiplying 7.20 mol \( Al \) by the conversion factor to find the moles of \( H_2SO_4 \) needed. 5. **Interactive Tools** - "ANSWER" space to enter the final calculation result. - "RESET" button to clear inputs and start over. **Conclusion:** Using stoichiometry and the balanced reaction, determine the exact amount of \( H_2SO_4 \) necessary for a complete reaction with aluminum, aiding in understanding chemical reaction equations and stoichiometric coefficients.
Transcribed Image Text:### Stoichiometry Problem: Reaction of Aluminum with Sulfuric Acid **Objective: Calculate Moles of \( H_2SO_4 \) Required** **Problem Statement:** What quantity of moles of \( H_2SO_4 \) is required to completely react with 7.20 moles of \( Al \), according to the balanced chemical equation: \[ 2 \, Al(s) + 3 \, H_2SO_4(aq) \rightarrow Al_2(SO_4)_3(aq) + 3 \, H_2(g) \] **Calculation Setup:** 1. **Starting Amount** - A box labeled "STARTING AMOUNT" indicates where to enter the initial quantity, in this case, 7.20 moles of \( Al \). 2. **Stoichiometric Conversion** - A visual setup for multiplication and addition factors helps to perform conversion from moles of \( Al \) to moles of \( H_2SO_4 \). \[ \begin{array}{|c|c|} \hline \text{Starting Moles} & \text{Conversion} \\ \hline \text{7.20 mol Al} & \left( \frac{3 \text{ mol } H_2SO_4}{2 \text{ mol } Al} \right) \\ \hline \end{array} \] 3. **Conversion Factors** - Various numeric buttons are available for factors such as 1, 98.08, 10.8, 26.98, and others, including Avogadro's number (6.022 × 10²³) for conversions. 4. **Result Calculation** - Complete the calculation by multiplying 7.20 mol \( Al \) by the conversion factor to find the moles of \( H_2SO_4 \) needed. 5. **Interactive Tools** - "ANSWER" space to enter the final calculation result. - "RESET" button to clear inputs and start over. **Conclusion:** Using stoichiometry and the balanced reaction, determine the exact amount of \( H_2SO_4 \) necessary for a complete reaction with aluminum, aiding in understanding chemical reaction equations and stoichiometric coefficients.
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