Question

3.8g of Zn reacts with 5.3g of CuCl2. if 1.6g of Cu is formed what is the percent yield of Cu?

Answer 1

no of moles of CuCl2 = weight of CuCl2 / molar mass of CuCl2

= 5.3 g / 134.45 g/mol

= 0.0394 moles

no of moles of Zn = weight of Zn / molar mass of Zn

= 3.8 g / 65.41 g/mol

= 0.058 moles

now write the balanced equation

Zn(s) + CuCl2(aq) --> ZnCl2(aq) + Cu(s)

from this equation it is clear that one mole of zinc and one mole of CuCl2 will produce one mole of Cu

but CuCl2 is having less moles than Zn

so limiting agent is CuCl2

so no of moles of Cu formed = no of moles of CuCl2

no of moles of Cu = 0.0394 moles this is theritically

theritical weight of Cu = moles of Cu x molar mass

= 0.0394 moles x 63.54 g /mole

= 2.5 grams

actually formed Cu = 1.6 grams given in the problem

now % of yield = (actual weight / theritical weight ) x 100

= (1.6 g / 2.5 g) x 100

= 64% yield