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14 October, 11:52

You are running a rather large scale reaction where you prepare the grignard reagent phenylmagnesium bromide by reacting 210.14 grams of magnesium with 772 ml of bromobenzene. How many moles of grignard reagent would you expect to form? (the density of bromobenzene is 1.495 g/ml, Mg = 24.3 g/mol, bromobenzene=157.01 g/mol)

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  1. 14 October, 14:23
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    We would expect to form 7.35 moles of grignard reagent.

    Explanation:

    Step 1: Data given

    Mass of magnesium = 210.14 grams

    Volume bromobenzene = 772 mL

    Density of bromobenzene = 1.495 g/mL

    Molar mass of Mg = 24.3 g/mol

    Molar mass of bromobenzene = 157.01 g/mol

    Step 2: The balanced equation

    C6H5Br + Mg ⇒ C6H5MgBr

    Step 3: Calculate mass of bromobenzene

    Mass bromobenzene = density bromobenzene * volume

    Mass bromobenzene = 1.495 g/mL * 772 mL

    Mass bromobenzene = 1154.14 grams

    Step 4: Calculate number of moles bromobenzene

    Moles bromobenzene = mass bromobenzene / molar mass bromobenzene

    Moles bromobenzene = 1154.14g / 157.01 g/mol

    Moles bromobenzene = 7.35 moles

    Step 5: Calculate moles of Mg

    Moles Mg = 210.14 grams / 24.3 g/mol

    Moles Mg = 8.65 moles

    Step 6: The limiting reactant

    The mole ratio is 1:1 So the bromobenzene has the smallest amount of moles, so it's the limiting reactant. It will be completely consumed (7.35 moles). Magnesium is in excess, There will react 7.35 moles. There will remain 8.65 - 7.35 = 1.30 moles

    Step 7: Calculate moles of phenylmagnesium bromide

    For 1 mole of bromobenzene, we need 1 mole of Mg to produce 1 mole of phenylmagnesium bromide

    For 7.35 moles bromobenzene, we have 7.35 moles phenylmagnesium bromide

    We would expect to form 7.35 moles of grignard reagent.
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