Aluminum reacts with bromine to form aluminum bromide: 2 Al + 3 Br2  2 AlBr3 if 48 grams of Al and 100 g of Br2 are reacted, how much AlBr3 can be produced

Md Ariful Islam

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E(Al)=48/(2*27)=0.889;

E(Br2)=100/(160*3)=0.208;

Br2 is the limitant reactant

m(AlBr3)=100*2*267/480=111.25 g;

Answer: 111.25 g of AlBr3 can be produced

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