Question

Aluminum will react with bromine to form aluminum bromide (Al₂Br₆). What mass of bromine (g) is needed to form 1.16 mol of Al₂Br₆?

a) 245.8 g
b) 98.2 g
c) 147.3 g
d) 73.6 g

Answer 1

To form 1.16 mol of aluminum bromide, approximately 277.7 g of bromine is needed.

Step-by-step explanation:

To calculate the mass of bromine needed to form 1.16 mol of aluminum bromide (Al₂Br₆), we need to use the balanced equation for the reaction:

2 Al (s) + 3 Br₂ (g) → Al₂Br₆ (s)

From the equation, we can see that 2 moles of aluminum bromide are formed for every 3 moles of bromine. Therefore, to form 1.16 mol of aluminum bromide, we need:

• (1.16 mol Al₂Br₆) × (3 mol Br₂ / 2 mol Al₂Br₆) = 1.74 mol Br2

To convert moles of bromine to grams, we can multiply the number of moles by the molar mass of bromine:

• (1.74 mol Br₂) × (159.808g / mol Br₂) = 277.7g

The mass of bromine needed to form 1.16 mol of aluminum bromide is approximately 277.7 g.