Question

3.0 g aluminum and 6.0 g of bromine react to form AlBr3 2Al+3Br2=2AlBr3 How much product would be produced? How much reagent would remain at the end of this reaction?

Answer 1

2Al + 3Br2 -------------> 2AlBr3

3 g Al = 0.11 mol Al. 

6 g Br2 = 0.0375 mole bromine (it is diatomic). 

moles of aluminium will take part in reaction = 0.0375 X (2/3) = 0.025. 

Gram-mole of AlBr3 will be produced = 0.025 mole = 6.6682 g. 
moles of Al left = 0.11 - 0.025 = 0.086