Question

Starting with the following equation, BCl₃(g) + LiAlH₄(s) → B₂H₆(g) + LiAlCl₄(s) calculate the moles of BCl₃ that will be required to produce 365 grams of B₂H₆.

a. 26.45 mol
b. 19.87 mol
c. 32.91 mol
d. 14.23 mol

Answer 1

The moles of BCl₃ required to produce 365 grams of B₂H₆ are calculated by first determining the molar mass of B₂H₆, converting the mass to moles, and then using the stoichiometry of the balanced chemical equation. Approximately 26.45 moles of BCl₃ are needed.

Step-by-step explanation:

To calculate the moles of BCl₃ required to produce 365 grams of B₂H₆, first determine the molar mass of B₂H₆ using the atomic masses from the periodic table. The molar mass of B₂H₆ is 10.81 g/mol (B) × 2 + 1.008 g/mol (H) × 6 = 27.66 g/mol. Then, using this molar mass, convert the given mass of B₂H₆ into moles:

365 g B₂H₆ × (1 mol B₂H₆ / 27.66 g B₂H₆) = 13.2 moles of B₂H₆

The balanced equation shows that 1 mole of BCl₃ gives 0.5 moles of B₂H₆, therefore:

13.2 moles B₂H₆ × (2 moles BCl₃ / 1 mole B₂H₆) = 26.4 moles BCl₃

Thus, approximately 26.45 moles of BCl₃ are needed to produce 365 grams of B₂H₆, which corresponds to answer choice (a).