Question

15mL of bromobenzene to grams to moles

6mL of nitric acid to g to moles

the limiting reactant of the two/ explain

theoretical yield of product

Answer 1

Answer –

We are given, volume of bromobenzene = 15 mL

, volume of nitric acid = 6 mL

Now first we need to calculate mass of bromobenzene –

We are given volume of bromobenzene and we know density of bromobenzene is 1.491 g/mL

So,

Density = mass / volume

Mass = density * volume

         = 1.491 g/mL * 15 mL

         = 22.365 g of bromobenzene

Now moles of bromobenzene –

Moles = given mass / molar mass of bromobenzene

Moles of bromobenzene = 22.365 g / 157.0 g.mol^-1

                                         = 0.142 moles of bromobenzene

Mass of nitric acid –

Density of nitric acid is 1.51 g/mL

So, mass of nitric acid = 6.0 mL * 1.51 g/mL

                                     = 9.06 g

Moles of nitric acid = 9.06 g / 63.01 g.mol^-1

                                  = 0.144 moles of nitric acid

We know reaction between bromobenzene and nitric acid and there is mole ratio is 1;1

C₆H₅Br + HNO₃ ----> C₆H₄BrNO₂ + HBr + H₂O

So all have mole ratio is 1:1

So moles of each reactant is equal to moles of product

So moles of C₆H₄BrNO₂ from C₆H₅Br is 0.142 moles

And moles of C₆H₄BrNO₂ from HNO₃ is 0.144 moles

So limiting reactant is bromobenzene and moles of C₆H₄BrNO₂ = 0.142 moles

So, theoretical yield of C₆H₄BrNO₂ = 0.142 moles * 202.0055 g/mole

                                                    = 28.8 g