4A1+30₂ -> 2Al₂O3 Given 6 moles of aluminum (Al), how many grams of aluminum oxide (Al2O3) will be produced? 152.97 grams Al₂O3 3 grams Al₂O3 305.94 grams Al2O3 11.34 grams Al2O3
4A1+30₂ -> 2Al₂O3 Given 6 moles of aluminum (Al), how many grams of aluminum oxide (Al2O3) will be produced? 152.97 grams Al₂O3 3 grams Al₂O3 305.94 grams Al2O3 11.34 grams Al2O3
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter3: Calculations With Chemical Formulas And Equaitons
Section: Chapter Questions
Problem 3.26QP
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Question
![---
### Stoichiometry Problem
#### Reaction Equation:
```
4Al + 3O₂ → 2Al₂O₃
```
#### Problem Statement:
Given 6 moles of aluminum (Al), how many grams of aluminum oxide (Al₂O₃) will be produced?
#### Options:
1. 152.97 grams Al₂O₃
2. 3 grams Al₂O₃
3. 305.94 grams Al₂O₃
4. 11.34 grams Al₂O₃
---
**Explanation:**
To solve this problem, follow these steps:
1. **Identify the stoichiometric ratio from the balanced equation:**
- 4 moles of Al produce 2 moles of Al₂O₃.
2. **Calculate the moles of Al₂O₃ produced from 6 moles of Al:**
- According to the stoichiometric ratio, \( \frac{4 \text{ moles of Al}}{2 \text{ moles of Al₂O₃}} = 2 \) moles of Al produce 1 mole of Al₂O₃.
- Therefore, six moles of Al will produce:
\[
6 \text{ moles of Al} \times \frac{2 \text{ moles of Al₂O₃}}{4 \text{ moles of Al}} = 3 \text{ moles of Al₂O₃}.
\]
3. **Calculate the mass of Al₂O₃ produced:**
- The molar mass of Al₂O₃ (\(\text{Al} = 27 \text{g/mol}, \text{O} = 16 \text{g/mol}\))
- Molar mass of Al₂O₃ =
\[
2(27 \text{ g/mol}) + 3(16 \text{ g/mol}) = 54 \text{ g/mol} + 48 \text{ g/mol} = 102 \text{ g/mol}.
\]
- So, the mass of 3 moles of Al₂O₃:
\[
3 \text{ moles} \times 102 \text{ g/mol} = 306 \text{ grams}.
\]
- Therefore, the correct answer is](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F062c447d-972d-42a3-bf9a-fbd1d95553cc%2F43c1fd59-08e8-4633-9919-e2cbf32e99d5%2Fy0m4mus_processed.jpeg&w=3840&q=75)
Transcribed Image Text:---
### Stoichiometry Problem
#### Reaction Equation:
```
4Al + 3O₂ → 2Al₂O₃
```
#### Problem Statement:
Given 6 moles of aluminum (Al), how many grams of aluminum oxide (Al₂O₃) will be produced?
#### Options:
1. 152.97 grams Al₂O₃
2. 3 grams Al₂O₃
3. 305.94 grams Al₂O₃
4. 11.34 grams Al₂O₃
---
**Explanation:**
To solve this problem, follow these steps:
1. **Identify the stoichiometric ratio from the balanced equation:**
- 4 moles of Al produce 2 moles of Al₂O₃.
2. **Calculate the moles of Al₂O₃ produced from 6 moles of Al:**
- According to the stoichiometric ratio, \( \frac{4 \text{ moles of Al}}{2 \text{ moles of Al₂O₃}} = 2 \) moles of Al produce 1 mole of Al₂O₃.
- Therefore, six moles of Al will produce:
\[
6 \text{ moles of Al} \times \frac{2 \text{ moles of Al₂O₃}}{4 \text{ moles of Al}} = 3 \text{ moles of Al₂O₃}.
\]
3. **Calculate the mass of Al₂O₃ produced:**
- The molar mass of Al₂O₃ (\(\text{Al} = 27 \text{g/mol}, \text{O} = 16 \text{g/mol}\))
- Molar mass of Al₂O₃ =
\[
2(27 \text{ g/mol}) + 3(16 \text{ g/mol}) = 54 \text{ g/mol} + 48 \text{ g/mol} = 102 \text{ g/mol}.
\]
- So, the mass of 3 moles of Al₂O₃:
\[
3 \text{ moles} \times 102 \text{ g/mol} = 306 \text{ grams}.
\]
- Therefore, the correct answer is
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