Answer:
Approximately 35.02 grams of H3BO3 will be formed when 6.11 grams of diborane (B2H6) reacts with excess water.
Step-by-step explanation:
To calculate the mass of H3BO3 formed when 6.11 g of diborane (B2H6) reacts with excess water, you need to first balance the chemical equation and then perform stoichiometric calculations.
The balanced chemical equation for the reaction is:
B2H6 + 6H2O → 2H3BO3 + 6H2
Now, we need to find the molar mass of B2H6 and H3BO3:
Molar mass of B2H6 (diborane):
B: 10.81 g/mol
H: 1.01 g/mol
Molar mass of B2H6 = (2 × 10.81 g/mol) + (6 × 1.01 g/mol) = 21.62 g/mol
Molar mass of H3BO3:
B: 10.81 g/mol
O: 16.00 g/mol
H: 1.01 g/mol
Molar mass of H3BO3 = (1 × 10.81 g/mol) + (3 × 1.01 g/mol) + (3 × 16.00 g/mol) = 61.84 g/mol
Now, let's calculate the number of moles of B2H6 in 6.11 g of B2H6:
Number of moles = Mass / Molar mass
Number of moles of B2H6 = 6.11 g / 21.62 g/mol ≈ 0.283 mol
According to the balanced equation, 1 mole of B2H6 produces 2 moles of H3BO3. So, the number of moles of H3BO3 produced will be:
Number of moles of H3BO3 = 0.283 mol × (2 moles H3BO3 / 1 mole B2H6) = 0.566 mol
Now, we can calculate the mass of H3BO3 formed:
Mass of H3BO3 = Number of moles × Molar mass
Mass of H3BO3 = 0.566 mol × 61.84 g/mol ≈ 35.02 g