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What mass of KBr is produced when 18.5 moles of K,SO, reacts? _AIBr3 +_K,SO4 → _KBr + _Al>(SO)3

What mass of KBr is produced when 18.5 moles of K,SO, reacts? _AIBr3 +_K,SO4 → _KBr + _Al>(SO)3

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
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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### Stoichiometry Problem: Determining Mass of KBr **Problem Statement:** What mass of KBr is produced when 18.5 moles of \( K_2SO_4 \) reacts? **Chemical Equation:** \[ \_\ AlBr_3 + \_\ K_2SO_4 \rightarrow \ KBr + \_\ Al_2(SO_4)_3 \] **Solution Explanation:** To solve this problem, follow these steps: 1. **Balance the Chemical Equation:** - Balance the reactants and products to ensure conservation of mass. 2. **Mole Ratio:** - Determine the mole ratio between \( K_2SO_4 \) and KBr. 3. **Calculate Moles of KBr:** - Use stoichiometry to convert moles of \( K_2SO_4 \) to moles of KBr. 4. **Find Mass of KBr:** - Convert moles of KBr to mass using the molar mass of KBr. Let's proceed through the solution step by step: 1. **Balancing the Equation:** To balance the equation, we determine the coefficients: \[ 2\ AlBr_3 + 3\ K_2SO_4 \rightarrow 6\ KBr + Al_2(SO_4)_3 \] 2. **Mole Ratio:** From the balanced equation, the mole ratio of \( K_2SO_4 \) to KBr is 3:6, which simplifies to 1:2. 3. **Using the Mole Ratio:** Given 18.5 moles of \( K_2SO_4 \): \[ 18.5\ moles\ K_2SO_4 \times \frac{6\ moles\ KBr}{3\ moles\ K_2SO_4} = 37\ moles\ KBr \] 4. **Converting to Mass:** The molar mass of KBr is approximately 119.0 g/mol. Therefore: \[ Mass\ of\ KBr = 37\ moles \times 119.0\ g/mol = 4403\ g\ KBr \] Thus, when 18.5 moles of \( K_2SO_4 \) react, 440
Transcribed Image Text:### Stoichiometry Problem: Determining Mass of KBr **Problem Statement:** What mass of KBr is produced when 18.5 moles of \( K_2SO_4 \) reacts? **Chemical Equation:** \[ \_\ AlBr_3 + \_\ K_2SO_4 \rightarrow \ KBr + \_\ Al_2(SO_4)_3 \] **Solution Explanation:** To solve this problem, follow these steps: 1. **Balance the Chemical Equation:** - Balance the reactants and products to ensure conservation of mass. 2. **Mole Ratio:** - Determine the mole ratio between \( K_2SO_4 \) and KBr. 3. **Calculate Moles of KBr:** - Use stoichiometry to convert moles of \( K_2SO_4 \) to moles of KBr. 4. **Find Mass of KBr:** - Convert moles of KBr to mass using the molar mass of KBr. Let's proceed through the solution step by step: 1. **Balancing the Equation:** To balance the equation, we determine the coefficients: \[ 2\ AlBr_3 + 3\ K_2SO_4 \rightarrow 6\ KBr + Al_2(SO_4)_3 \] 2. **Mole Ratio:** From the balanced equation, the mole ratio of \( K_2SO_4 \) to KBr is 3:6, which simplifies to 1:2. 3. **Using the Mole Ratio:** Given 18.5 moles of \( K_2SO_4 \): \[ 18.5\ moles\ K_2SO_4 \times \frac{6\ moles\ KBr}{3\ moles\ K_2SO_4} = 37\ moles\ KBr \] 4. **Converting to Mass:** The molar mass of KBr is approximately 119.0 g/mol. Therefore: \[ Mass\ of\ KBr = 37\ moles \times 119.0\ g/mol = 4403\ g\ KBr \] Thus, when 18.5 moles of \( K_2SO_4 \) react, 440
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