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

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

Hg was the limiting reagent, not Br2.

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

To determine the number of grams of HgBr2 that can be produced, we need to use stoichiometry.

First, let's write the balanced chemical equation for the reaction between Hg and Br2:
Hg + Br2 -> HgBr2

The molar mass of Hg is 200.59 g/mol, and the molar mass of Br2 is 159.808 g/mol.

Step 1: Convert the given masses of Hg and Br2 to moles.
Moles of Hg = mass of Hg / molar mass of Hg
Moles of Hg = 10.0 g / 200.59 g/mol
Moles of Hg = 0.0498 mol

Moles of Br2 = mass of Br2 / molar mass of Br2
Moles of Br2 = 10 g / 159.808 g/mol
Moles of Br2 = 0.0626 mol

Step 2: Use the stoichiometry from the balanced equation to determine the moles of HgBr2 produced.
From the balanced equation: 1 mol Hg reacts with 1 mol Br2 to form 1 mol HgBr2.

Since the stoichiometry is 1:1:1, the number of moles of HgBr2 formed will be the same as the number of moles of Hg or Br2 in the reaction.

Therefore, the moles of HgBr2 produced = 0.0498 mol (or 0.0626 mol, since they are the same)

Step 3: Convert moles of HgBr2 to grams.
Mass of HgBr2 = moles of HgBr2 × molar mass of HgBr2
Mass of HgBr2 = 0.0498 mol × (HgBr2 molar mass)

The molar mass of HgBr2 can be calculated by adding the molar masses of Hg (200.59 g/mol) and Br2 (159.808 g/mol):
Molar mass of HgBr2 = 200.59 g/mol + 2 × 79.904 g/mol
Molar mass of HgBr2 = 360.398 g/mol

Mass of HgBr2 = 0.0498 mol × 360.398 g/mol
Mass of HgBr2 = 17.954 g

Therefore, approximately 17.954 grams of HgBr2 can be produced when 10.0 grams of Hg reacts with 10 grams of Br2.