Question

Calculate the number of grams of bromine (Br₂) needed to react exactly with 50.1 g aluminium (Al). The equation for the reaction is:

2Al (s) + 3Br₂ (l) → 2AlBr₃ (s)

Then, calculate the number of moles of AlBr₃ produced by the process.

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

 

Moles of Al $=\frac{50.1}{27}=1.85 moles$

Moles of Br₂$=\frac {3}{2}×1.85=2.775moles$

Mass of Br₂ $=2.775×159.8=443.45grams$

Moles of AlBr₃$= 1.85$