Question

What mass of liquid C6H13Br will be produced by a reaction giving 85% yield if 17.0 mL of a 3.0M solution of C6H12 is treated with 3.10 L HBr(g) at STP?

Answer 1

Moles of C6H12 = molarity x volume

= 3 mol/L x 17 mL x 1L/1000 mL

= 0.051 mol

Moles of HBr = PV/RT

= 1 atm x 3.10 L / 0.0821 L-atm/mol-K x 273K

= 0.1383 mol

The balanced reaction

C6H12 + HBr = C6H13Br

1 mol C6H12 required = 1 mol HBr

0.051 mol C6H12 required = 0.051 mol HBr

But we have more moles than required

Limiting reactant = C6H12

Moles of C6H13Br produced = 0.051 mol

Theoretical yield of C6H13Br = moles x molecular weight

= 0.051 mol x 165.07 g/mol

= 8.4185 g

% yield = actual yield x 100 / theoretical yield

85/100 = actual yield / 8.4185

actual yield = 7.156 g