Initial mass of triphenyl methanol in g = 0.220g Initial volume of 33% HBr solution in ml = 0.60 ml Find mas of triphenyl bromide in g = 240 g

0.792g of triphenyl bromide are produced.

Explanation:

The reaction of triphenyl methanol with HBr is:

triphenyl methanol + HBr → Triphenyl bromide.

Reaction (1:1), 1 mole of HBr reacts per mole of triphenyl methanol.

To know the mass of triphenyl bromide assuming a theoretical yield (Yield 100%) we need to find first limiting reactant:

Moles triphenyl methanol (Molar mass: 260.33g/mol) =

0.220g × (1mol / 260.33g) = 8.45x10⁻³ moles Triphenyl methanol

Moles HBr (Molar mass: 80.91g/mol; 33%=33g HBr/100mL) =

0.60mL ₓ (33g / 100mL) ₓ (1mol / 80.91g) = 2.45x10⁻³ moles HBr

As amount of moles of HBr is lower than moles of triphenyl methanol, HBr is limiting reactant.

As HBr is limiting reactant, moles produced of triphenyl bromide = moles HBr = 2.45x10⁻³ moles

As molar mass of triphenyl bromide is 323.2g/mol, mass of triphenyl bromide is:

2.45x10⁻³ moles × (323.2g / mol) =

i believe it is 40–40–20%

explanation: