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?

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

The answer to that would be magnesium