Answer :
Let's solve the problem step by step:
### Step 1: Calculate the moles of Oâ‚‚ gas
At Standard Temperature and Pressure (STP), 1 mole of any gas occupies 22.4 liters. Given that we have [tex]\( 3.25 \)[/tex] liters of Oâ‚‚, we can find the number of moles of Oâ‚‚ as follows:
[tex]\[ \text{Moles of Oâ‚‚} = \frac{3.25 \text{ liters}}{22.4 \text{ liters/mole}} \][/tex]
[tex]\[ \text{Moles of Oâ‚‚} = 0.14508928571428573 \text{ moles} \][/tex]
### Step 2: Use stoichiometry to find moles of B₂O₃
From the balanced chemical equation:
[tex]\[ 2 \text{ B}_5\text{H}_9 + 12 \text{ O}_2 \rightarrow 5 \text{ B}_2\text{O}_3 + 9 \text{ H}_2\text{O} \][/tex]
We see that 12 moles of O₂ produce 5 moles of B₂O₃. The molar ratio of O₂ to B₂O₃ is [tex]\(\frac{5}{12}\)[/tex].
Using the moles of Oâ‚‚ we calculated:
[tex]\[ \text{Moles of B₂O₃} = \text{Moles of O₂} \times \frac{5}{12} \][/tex]
[tex]\[ \text{Moles of B₂O₃} = 0.14508928571428573 \times \frac{5}{12} \][/tex]
[tex]\[ \text{Moles of B₂O₃} = 0.060453869047619055 \text{ moles} \][/tex]
### Step 3: Calculate the mass of B₂O₃ produced
To find the grams of B₂O₃, we use its molar mass. The molar mass of B₂O₃ is calculated as follows:
- Boron (B) has an atomic mass of 10.81 g/mole.
- Oxygen (O) has an atomic mass of 16 g/mole.
The molar mass of B₂O₃:
[tex]\[ (2 \times 10.81) + (3 \times 16) = 21.62 + 48 = 69.62 \text{ g/mole} \][/tex]
Using the moles of B₂O₃ we calculated:
[tex]\[ \text{Mass of B₂O₃} = \text{Moles of B₂O₃} \times \text{Molar mass of B₂O₃} \][/tex]
[tex]\[ \text{Mass of B₂O₃} = 0.060453869047619055 \times 69.62 \][/tex]
[tex]\[ \text{Mass of B₂O₃} = 4.208798363095239 \text{ grams} \][/tex]
Thus, the mass of B₂O₃ produced is approximately 4.209 grams.
### Step 1: Calculate the moles of Oâ‚‚ gas
At Standard Temperature and Pressure (STP), 1 mole of any gas occupies 22.4 liters. Given that we have [tex]\( 3.25 \)[/tex] liters of Oâ‚‚, we can find the number of moles of Oâ‚‚ as follows:
[tex]\[ \text{Moles of Oâ‚‚} = \frac{3.25 \text{ liters}}{22.4 \text{ liters/mole}} \][/tex]
[tex]\[ \text{Moles of Oâ‚‚} = 0.14508928571428573 \text{ moles} \][/tex]
### Step 2: Use stoichiometry to find moles of B₂O₃
From the balanced chemical equation:
[tex]\[ 2 \text{ B}_5\text{H}_9 + 12 \text{ O}_2 \rightarrow 5 \text{ B}_2\text{O}_3 + 9 \text{ H}_2\text{O} \][/tex]
We see that 12 moles of O₂ produce 5 moles of B₂O₃. The molar ratio of O₂ to B₂O₃ is [tex]\(\frac{5}{12}\)[/tex].
Using the moles of Oâ‚‚ we calculated:
[tex]\[ \text{Moles of B₂O₃} = \text{Moles of O₂} \times \frac{5}{12} \][/tex]
[tex]\[ \text{Moles of B₂O₃} = 0.14508928571428573 \times \frac{5}{12} \][/tex]
[tex]\[ \text{Moles of B₂O₃} = 0.060453869047619055 \text{ moles} \][/tex]
### Step 3: Calculate the mass of B₂O₃ produced
To find the grams of B₂O₃, we use its molar mass. The molar mass of B₂O₃ is calculated as follows:
- Boron (B) has an atomic mass of 10.81 g/mole.
- Oxygen (O) has an atomic mass of 16 g/mole.
The molar mass of B₂O₃:
[tex]\[ (2 \times 10.81) + (3 \times 16) = 21.62 + 48 = 69.62 \text{ g/mole} \][/tex]
Using the moles of B₂O₃ we calculated:
[tex]\[ \text{Mass of B₂O₃} = \text{Moles of B₂O₃} \times \text{Molar mass of B₂O₃} \][/tex]
[tex]\[ \text{Mass of B₂O₃} = 0.060453869047619055 \times 69.62 \][/tex]
[tex]\[ \text{Mass of B₂O₃} = 4.208798363095239 \text{ grams} \][/tex]
Thus, the mass of B₂O₃ produced is approximately 4.209 grams.
