Asked by Chris
How many grams of HgBr2 can be produced if 10.0 grams of Hg reacts with 10 grams Br2
Answers
Answered by
Devron
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
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
Answered by
Devron
Hg was the limiting reagent, not Br2.
moles of Hg/360.41 g of HgBr2/mol= g of HgBr2
moles of Hg/360.41 g of HgBr2/mol= g of HgBr2
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