Question

Aluminum reacts with bromine, producing aluminum bromide. In a certain experiment, 20 ml of bromine was used. What mass of aluminum bromide would be produced assuming complete reaction?

a) Depends on other factors
b) 202.6 g
c) 160.6 g
d) 243.2 g

Answer 1

The mass of aluminum bromide produced can be calculated by converting the volume of bromine used to moles and then using stoichiometry. The correct answer is 243.2 g.

Step-by-step explanation:

In the reaction between aluminum and bromine, aluminum reacts with bromine to produce aluminum bromide. The balanced equation for this reaction is:

2 Al (s) + 3 Br2 (g) → 2 AlBr3 (s)

To determine the mass of aluminum bromide produced, we need to calculate the molar mass of aluminum bromide and then use stoichiometry to convert the volume of bromine used to mass of aluminum bromide.

The molar mass of aluminum bromide (AlBr3) is 266.7 g/mol. Since the molar ratio between aluminum and aluminum bromide is 2:2, the moles of aluminum bromide produced will be equal to the moles of aluminum used. To find the moles of aluminum, we need to use the volume of bromine used and its molar volume at standard conditions (22.4 L/mol):

20 mL bromine × (1 L/1000 mL) × (1 mol/22.4 L) = 0.0008929 mol

Therefore, the mass of aluminum bromide produced is:

0.0008929 mol × 266.7 g/mol = 0.2385 g

The correct answer is d) 243.2 g.