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8 November, 21:50

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?

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  1. 8 November, 22:11
    0
    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
  2. 8 November, 22:18
    0
    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.
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