Answer to Question #122943 in Inorganic Chemistry for lia

Question

if 48 grams of Al and 100 g of Br2 are reacted, how much AlBr3 can be produced

Answer 1

2Al+3Br₂=2AlBr₃

1.we can find out which substance is left over.for this we first find the amount of Al and Br₂ in (mol)

n_(Al)=48/27=1.778mol

n_(Br2)=100/160=0.625mol

2.according to the above calculations, Al increases. Therefore, we can find AlBr₃ based on the amount (mol) of Br_2:

3mol Br₂——> 2 mol AlBr₃

0.625mol Br₂——> x mol AlBr₃

x=0.625*2/3=0.4167 mol AlBr₃

3.now we find the mass of AlBr3. for this we use the following formula: m = n * M

m=0.4167*267=111.25 gr AlBr₃

Answer: 111.25 gr