Mercury and bromine willreact with each other to produce mercury (II) bromide:Hg (l) + Br2 (l) - ->HgBr2 (s) Consider an experimentwhere 10.0 g Hg is reacted with 9.00 g Br2. (a) What is the limitingreactant under these conditions? (b) What mass ofHgBr2 canbe produced from this reaction under these conditions? (c) How many moles ofwhich reactant will be left unreacted? What mass of that reactant is left unreacted?

a. Limiting reagent is the Hg.

b. 17.9 g of HgBr₂ will be produced in the reaction.

c. 6.5x10⁻³ moles of Br₂, are without reaction and the mass for that, is 1.03 g.

Explanation:

This is the reaction:

Hg (l) + Br₂ (l) - -> HgBr₂ (s)

Ratio is 1:1, so 1 mol of mercury reacts with 1 mol of bromine.

Mass / Molar mass = Mol

10 g / 200.59 g/m = 0.0498 moles of Hg

9 g / 159.8 g/m = 0.0563 moles of Br₂

As ratio is 1:1, and for 0.0563 moles of Br₂ I need the same amount of Hg to react, I only have just 0.0498 so my limiting reactant is the Hg.

Ratio with products is also 1:1, so 1 mol of Hg makes 1 mol of bromide.

Then 0.0498 moles of Hg will produce 0.0498 moles of bromide.

Molar mass = 360.39 g/m

Mol. molar mass = mass → 0.0498 m. 360.39 g/m = 17.9 g of HgBr₂

To react all the Br₂ I have to consume the 0.0563 moles of it, but I only have 0.0498 moles for reacting, so the moles that will be left unreacted are:

0.0563 m - 0.0498 m = 6.5x10⁻³ moles of Br₂ (without reaction)

Mol. molar mass = mass → 6.5x10⁻³ mol. 159 g/m = 1.03 grams without reaction.

a) The Hg is the limiting reactant

b) There is 17.98 grams of HgBr2 produced

c) There will rmain 0.0064 moles of Br2 unreacted, this is 1.02 grams

Explanation:

Step 1: Data given

Mass of Hg = 10.0 grams

Mass of Br2 = 9.00 grams

Molar mass of Hg = 200.59 g/mol

Molar mass of Br2 = 159.8 g/mol

Step 2: The balanced equation

Hg (l) + Br2 (l) → HgBr2 (s)

Step 3: Calculate moles Hg

Moles Hg = mass Hg / molar mass Hg

Moles Hg = 10.0 grams / 200.59 g/mol

Moles Hg = 0.0499 moles

Step 4: Calculate moles Br2

Moles Br2 = 9.00 grams / 159.8 g/mol

Moles Br2 = 0.0563 moles

Step 5: Calculate the limiting reactant

For 1 mol Hg we need 1 mol Br2 to produce 1 mol of HgBr2

Hg is the limiting reactant. It will completely be consumed (0.0499 moles)

Br2 is in excess. There will remain 0.0563 - 0.0499 = 0.0064 moles

This is 0.0064 moles * 159.8 = 1.02 grams

Step 6: Calculate moles HgBr2

For 1 mol Hg we need Br2 to produce 1 mol HgBr2

For 0.0499 moles Hg we'll have 0.0499 moles HgBr2

Step 7: Calculate mass of HgBr2

Mass HgBr2 = 0.0499 moles * 360.4 g/mol

Mass HgBr2 = 17.98 grams