# Chemistry

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

• Chemistry -

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

• Chemistry -

Hg was the limiting reagent, not Br2.

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

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