Question
From the reaction: B2H6 +3O2 ---> 2HBO2 +2H2O
a. What mass of O2 will be needed to burn 36.1 g of B2H6?
b. How many moles of water are produced from 19.2 g of B2H6?
a. What mass of O2 will be needed to burn 36.1 g of B2H6?
b. How many moles of water are produced from 19.2 g of B2H6?
Answers
GPT-4o mini
To solve the problems, we first need to understand the stoichiometry of the reaction. The balanced equation given is:
\[
\text{B}_2\text{H}_6 + 3\text{O}_2 \rightarrow 2\text{HBO}_2 + 2\text{H}_2\text{O}
\]
### Part (a): Mass of \( \text{O}_2 \) needed to burn 36.1 g of \( \text{B}_2\text{H}_6 \)
1. **Calculate the molar mass of \( \text{B}_2\text{H}_6 \)**:
- Boron (B): 10.81 g/mol
- Hydrogen (H): 1.01 g/mol
\[
\text{Molar mass of } \text{B}_2\text{H}_6 = (2 \times 10.81) + (6 \times 1.01) = 21.62 + 6.06 = 27.68 \text{ g/mol}
\]
2. **Determine the number of moles of \( \text{B}_2\text{H}_6 \) in 36.1 g**:
\[
\text{Moles of } \text{B}_2\text{H}_6 = \frac{36.1 \text{ g}}{27.68 \text{ g/mol}} \approx 1.30 \text{ moles}
\]
3. **Use the stoichiometry of the reaction to find the moles of \( \text{O}_2 \)**:
- According to the balanced equation, 1 mole of \( \text{B}_2\text{H}_6 \) reacts with 3 moles of \( \text{O}_2 \).
- Therefore, \( 1.30 \text{ moles of } \text{B}_2\text{H}_6 \) will require:
\[
\text{Moles of } \text{O}_2 = 1.30 \text{ moles } \times 3 = 3.90 \text{ moles of } \text{O}_2
\]
4. **Convert moles of \( \text{O}_2 \) to mass**:
\[
\text{Molar mass of } \text{O}_2 = 2 \times 16.00 = 32.00 \text{ g/mol}
\]
\[
\text{Mass of } \text{O}_2 = 3.90 \text{ moles} \times 32.00 \text{ g/mol} = 124.8 \text{ g}
\]
### Part (b): Moles of water produced from 19.2 g of \( \text{B}_2\text{H}_6 \)
1. **Calculate the number of moles of \( \text{B}_2\text{H}_6 \) in 19.2 g** (using the molar mass calculated earlier):
\[
\text{Moles of } \text{B}_2\text{H}_6 = \frac{19.2 \text{ g}}{27.68 \text{ g/mol}} \approx 0.693 \text{ moles}
\]
2. **Use the stoichiometry of the reaction to find the moles of \( \text{H}_2\text{O} \)**:
- The balanced equation shows that 1 mole of \( \text{B}_2\text{H}_6 \) produces 2 moles of \( \text{H}_2\text{O} \). Therefore,
\[
\text{Moles of } \text{H}_2\text{O} = 0.693 \text{ moles of } \text{B}_2\text{H}_6 \times 2 = 1.386 \text{ moles of } \text{H}_2\text{O}
\]
### Answers:
a. The mass of \( \text{O}_2 \) needed to burn 36.1 g of \( \text{B}_2\text{H}_6 \) is approximately 124.8 g.
b. The number of moles of \( \text{H}_2\text{O} \) produced from 19.2 g of \( \text{B}_2\text{H}_6 \) is approximately 1.386 moles.
\[
\text{B}_2\text{H}_6 + 3\text{O}_2 \rightarrow 2\text{HBO}_2 + 2\text{H}_2\text{O}
\]
### Part (a): Mass of \( \text{O}_2 \) needed to burn 36.1 g of \( \text{B}_2\text{H}_6 \)
1. **Calculate the molar mass of \( \text{B}_2\text{H}_6 \)**:
- Boron (B): 10.81 g/mol
- Hydrogen (H): 1.01 g/mol
\[
\text{Molar mass of } \text{B}_2\text{H}_6 = (2 \times 10.81) + (6 \times 1.01) = 21.62 + 6.06 = 27.68 \text{ g/mol}
\]
2. **Determine the number of moles of \( \text{B}_2\text{H}_6 \) in 36.1 g**:
\[
\text{Moles of } \text{B}_2\text{H}_6 = \frac{36.1 \text{ g}}{27.68 \text{ g/mol}} \approx 1.30 \text{ moles}
\]
3. **Use the stoichiometry of the reaction to find the moles of \( \text{O}_2 \)**:
- According to the balanced equation, 1 mole of \( \text{B}_2\text{H}_6 \) reacts with 3 moles of \( \text{O}_2 \).
- Therefore, \( 1.30 \text{ moles of } \text{B}_2\text{H}_6 \) will require:
\[
\text{Moles of } \text{O}_2 = 1.30 \text{ moles } \times 3 = 3.90 \text{ moles of } \text{O}_2
\]
4. **Convert moles of \( \text{O}_2 \) to mass**:
\[
\text{Molar mass of } \text{O}_2 = 2 \times 16.00 = 32.00 \text{ g/mol}
\]
\[
\text{Mass of } \text{O}_2 = 3.90 \text{ moles} \times 32.00 \text{ g/mol} = 124.8 \text{ g}
\]
### Part (b): Moles of water produced from 19.2 g of \( \text{B}_2\text{H}_6 \)
1. **Calculate the number of moles of \( \text{B}_2\text{H}_6 \) in 19.2 g** (using the molar mass calculated earlier):
\[
\text{Moles of } \text{B}_2\text{H}_6 = \frac{19.2 \text{ g}}{27.68 \text{ g/mol}} \approx 0.693 \text{ moles}
\]
2. **Use the stoichiometry of the reaction to find the moles of \( \text{H}_2\text{O} \)**:
- The balanced equation shows that 1 mole of \( \text{B}_2\text{H}_6 \) produces 2 moles of \( \text{H}_2\text{O} \). Therefore,
\[
\text{Moles of } \text{H}_2\text{O} = 0.693 \text{ moles of } \text{B}_2\text{H}_6 \times 2 = 1.386 \text{ moles of } \text{H}_2\text{O}
\]
### Answers:
a. The mass of \( \text{O}_2 \) needed to burn 36.1 g of \( \text{B}_2\text{H}_6 \) is approximately 124.8 g.
b. The number of moles of \( \text{H}_2\text{O} \) produced from 19.2 g of \( \text{B}_2\text{H}_6 \) is approximately 1.386 moles.