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)

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.

h let me think