If you react 45.0 g of Br2 with 30.0 g of Al, how many grams of product would you expect to make? 2Al(s)+ 3Br2(l) —> 2AlBr3(s)

Question

How many grams would you expect to make?

Answer 1

As per the chemical equation

2 moles of Aluminum react with 3 moles of Bromine to generate 2 moles of Aluminum Bromide

 

Molecular Weight of Al= 27 g/mol

Molecular Weight of Bromine= 159.8 g/mol

Molecular Weight of Aluminum Bromide= 266.7 g/mol

 

Therefore, 2 moles of Aluminum= 54 g

3 moles of Bromine= 479.4 g

2 moles of Aluminum Bromide= 533.4 g

 

As Bromine is needed in the largest amount for reactants it is the limiting amount.

Therefore, 45 g of Bromine will react completely to generate (533.4/479.4)*45 g of AlBr3

 

Total grams made= 50.1 g