Calculate the mass of aluminum bromide which is formed by the interaction of 5.4 g of aluminum with an excess of bromine.

Given:

m (Al) = 5.4 g

To find:

m (AlBr3) -?

Solution:

2Al + 3Br2 = 2AlBr3, – we solve the problem, relying on the composed reaction equation:

1) Find the amount of aluminum that has reacted:

n (Al) = m: M = 5.4 g: 27 g / mol = 0.2 mol

2) We compose a logical expression:

If 2 mol of Al gives 2 mol of AlBr3,

then 0.2 mol of Al will give x mol of AlBr3,

then x = 0.2 mol.

3) Find the mass of aluminum bromide formed during the reaction:

m (AlBr3) = n * M = 0.2 mol * 267 g / mol = 53.4 g.

Answer: m (AlBr3) = 53.4 g.