Mercury and bromine react with each other to produce mercury to bromid Mercury and bro mine will react with each other to produce mercury(ll) bromide

Question

Hg(l)+Br2(l)——->HgBr2(s)

a) what mass of HgBr2 can be produced from the reaction of 10g Hg and 9g of Br2

Answer 1

Balanced equation is Hg(l)+Br2(l)——->HgBr2(s)

moles Hg used = 10 g x 1 mole/200.6 g = 0.04985 moles

moles Br₂ used = 9 g x 1 mol/159.8 g = 0.0563 mokes

Limiting reactant is Hg

moles HBr₂ formed = 0.04985 mol Hg x 1 mol HgBr₂/mol Hg = 0.04985 moles HBr₂

mass HBr₂ = 0.04985 mol x 360 g/mol = 17.946 g = 20 g (to 1 sig fig)