Final answer:
To produce 735 grams of B₂H₆, we calculate that 53.12 moles of BCl₃ are required based on the molar mass of B₂H₆ and the stoichiometry of the given chemical equation.
Step-by-step explanation:
To calculate the moles of BCl₃ that will be required to produce 735 grams of B₂H₆, we first need to determine the molar mass of B₂H₆. The molar mass of B₂H₆ is 27.67 g/mol (10.81 g/mol for boron x 2 + 1.01 g/mol for hydrogen x 6). Now, we use this to find out how many moles are in 735 grams of B₂H₆:
735 g B₂H₆ × (1 mol B₂H₆ / 27.67 g B₂H₆) = 26.56 mol B₂H₆
From the balanced chemical equation, it can be seen that 1 mole of BCl₃ produces 0.5 moles of B₂H₆. Therefore, the number of moles of BCl₃ required is double the number of moles of B₂H₆.
26.56 mol B₂H₆ × (2 mol BCl₃ / 1 mol B₂H₆) = 53.12 mol BCl₃
Therefore, 53.12 moles of BCl₃ are required to produce 735 grams of B₂H₆.