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(SHOW CALCULATIONS) What mass of Al(s) can be formed from 20.0 g of Al2O3(s) and 50.0 g of C(s) by the following reaction? Al2O3(s) + 3C(s) 2Al(s) + 3CO(g)

(SHOW CALCULATIONS) What mass of Al(s) can be formed from 20.0 g of Al2O3(s) and 50.0 g of C(s) by the following reaction? Al2O3(s) + 3C(s) 2Al(s) + 3CO(g)

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 Statement What mass of \( \text{Al}(s) \) can be formed from 20.0 g of \( \text{Al}_2\text{O}_3(s) \) and 50.0 g of \( \text{C}(s) \) by the following reaction? \[ \text{Al}_2\text{O}_3(s) + 3\text{C}(s) \rightarrow 2\text{Al}(s) + 3\text{CO}(g) \] ### Steps to Solve the Problem 1. **Calculate the molar masses of the reactants and products:** - Molar mass of \( \text{Al}_2\text{O}_3(s) \): \[ 2 \times 27.0\ \text{g/mol (Al)} + 3 \times 16.0\ \text{g/mol (O)} = 102.0\ \text{g/mol} \] - Molar mass of \( \text{C}(s) \): \[ 12.0\ \text{g/mol} \] - Molar mass of \( \text{Al}(s) \): \[ 27.0\ \text{g/mol} \] 2. **Convert the given masses to moles:** - Moles of \( \text{Al}_2\text{O}_3(s) \): \[ \frac{20.0\ \text{g}}{102.0\ \text{g/mol}} = 0.196\ \text{mol} \] - Moles of \( \text{C}(s) \): \[ \frac{50.0\ \text{g}}{12.0\ \text{g/mol}} = 4.167\ \text{mol} \] 3. **Determine the limiting reactant:** - Based on the stoichiometry of the reaction, 1 mole of \( \text{Al}_2\text{O}_3 \) reacts with 3 moles of \( \text{C} \). - Moles of \( \text{C} \) needed to react with 0.196 moles of \( \text{Al}_2\text{O}_3 \
Transcribed Image Text:### Problem Statement What mass of \( \text{Al}(s) \) can be formed from 20.0 g of \( \text{Al}_2\text{O}_3(s) \) and 50.0 g of \( \text{C}(s) \) by the following reaction? \[ \text{Al}_2\text{O}_3(s) + 3\text{C}(s) \rightarrow 2\text{Al}(s) + 3\text{CO}(g) \] ### Steps to Solve the Problem 1. **Calculate the molar masses of the reactants and products:** - Molar mass of \( \text{Al}_2\text{O}_3(s) \): \[ 2 \times 27.0\ \text{g/mol (Al)} + 3 \times 16.0\ \text{g/mol (O)} = 102.0\ \text{g/mol} \] - Molar mass of \( \text{C}(s) \): \[ 12.0\ \text{g/mol} \] - Molar mass of \( \text{Al}(s) \): \[ 27.0\ \text{g/mol} \] 2. **Convert the given masses to moles:** - Moles of \( \text{Al}_2\text{O}_3(s) \): \[ \frac{20.0\ \text{g}}{102.0\ \text{g/mol}} = 0.196\ \text{mol} \] - Moles of \( \text{C}(s) \): \[ \frac{50.0\ \text{g}}{12.0\ \text{g/mol}} = 4.167\ \text{mol} \] 3. **Determine the limiting reactant:** - Based on the stoichiometry of the reaction, 1 mole of \( \text{Al}_2\text{O}_3 \) reacts with 3 moles of \( \text{C} \). - Moles of \( \text{C} \) needed to react with 0.196 moles of \( \text{Al}_2\text{O}_3 \
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