Stoichiometry question + excess and mass?

Question

I have the question...

Hg (l) + Br2 (l) --> HgBr2 (s)

2.15g of Hg is present with 1.56g of Br2

I found that the mass of HgBr2 produced was 3.52g

And the question is now asking me to find the mass of the reactant that remains in excess.

How do I do that?

Answer 1

You had 0.01075 moles of Hg and 0.00975 moles of Br2.  One mole of Hg reacts with one mole of Br2, so Br2 is the limiting reagent (there are too many moles of Hg). The resulting HgBr2 is 0.00975*(260) g, which gives your answer.  However, 0.01075 - 0.00975 moles of Hg remain unreacted, so the mass of the excess is (0.01075 - 0.00975)*200.  (I used 200 for atomic mass of Hg and 160 for molecular mass of Br2)