A chemist mixed 3.13 g of aluminium with 28.8 g of bromine to form aluminium bromide according to the equation

A chemist mixed 3.13 g of aluminium with 28.8 g of bromine to form aluminium bromide according to the equation: 2Al + 3Br2 -> 2AlBr3

(a) Identify the limiting reagent. (please enter chemical symbol)
(b) What is the maximum mass of aluminium bromide which can be made? g
(c) With respect to the reactant that is in excess, what mass remains after the reaction is completed? g Br2 remains

I answered
a) Al is limiting
b) 32.0 grams
c) 0.991 grams
I was wrong, can you please help me solve this question.

Moles of Al = 3.13/27 = 0.116 mol
Moles of Br2 = 28.8/156 = 0.185 mol

0.086 moles of Al need, 0.116/2*3 = 0.174mol Br2 to completely react.
Since this is there in excess, Br2 is in excess and Al is the limiting factor

All Al will react, forming 0.116/2*2 =0.116 mol of AlBr3
Mass of AlBr3, 0.116mol * 266.7 = 30.9 g

Br2 is in excess, 0.185mol - 0.174mol = 0.011 mol
= 0.011*156 = 1.716g

No probs, always ready to help:)

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