How many grams of HgBr2 can be produced if 10.0 grams of Hg reacts with 10 grams Br2

Answer 1

You need to first find the limiting reagent/element in the formula. Solve for moles for each.

10.0g of Hg*(1 mole of Hg/200.6g of Hg)= moles of Hg

10.0g of Br2*(1 mole of Br2/159.8g of Br)= moles of Br

Looking at it Br2 will be your limiting element/reagent

moles of Br2/360.41 g of HgBr2/mol= g of HgBr2

Answer 2

Hg was the limiting reagent, not Br2.

moles of Hg/360.41 g of HgBr2/mol= g of HgBr2