Answer:
We would expect to form 7.35 moles of grignard reagent.
Explanation:
<u>Step 1: </u>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
<u>Step 2</u>: The balanced equation
C6H5Br + Mg ⇒ C6H5MgBr
<u>Step 3:</u> Calculate mass of bromobenzene
Mass bromobenzene = density bromobenzene * volume
Mass bromobenzene = 1.495 g/mL * 772 mL
Mass bromobenzene = 1154.14 grams
<u>Step 4</u>: 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
<u>Step 5:</u> Calculate moles of Mg
Moles Mg = 210.14 grams /24.3 g/mol
Moles Mg = 8.65 moles
<u>Step 6:</u> 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
<u>Step 7:</u> 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.
