Question

Aluminum reacts with bromine to form aluminum bromide (used as an acid catalyst in organic synthesis). Al(s) + Br₂(l) → Al₂Br₆(s) [unbalanced] How many moles of Al are needed to form 2.43 mol of Al₂Br₆?

A. 1.62 mol
B. 7.29 mol
C. 1.22 mol
D. 2.43 mol
E. 4.86 mol

Answer 1

To form 2.43 mol of aluminum bromide (Al₂Br₆), 2.43 moles of aluminum (Al) are needed according to the balanced chemical reaction. Thus, the correct answer is option D. 2.43 mol.

Step-by-step explanation:

The student has asked about the stoichiometry of the reaction between aluminum (Al) and bromine (Br₂) to form aluminum bromide (Al₂Br₆). To find out how many moles of Al are needed to form 2.43 mol of Al₂Br₆, we need to balance the reaction equation and use the stoichiometric ratios. The balanced equation is 2 Al (s) + 3 Br₂ (l) → 2 Al₂Br₆ (s).

From the balanced equation, it can be seen that 2 moles of Al react to form 2 moles of Al₂Br₆. Therefore, for every mole of Al₂Br₆ produced, 1 mole of Al is needed. To produce 2.43 mol of Al₂Br₆, the amount of Al required is:

2.43 mol Al₂Br₆ × (1 mol Al / 1 mol Al₂Br₆) = 2.43 mol Al

The direct answer to the student's question is option D. 2.43 mol.