Question

If 1.357 g of cadmium sulfate is mixed with 0.7519 g of potassium sulfide, what is the percent yield of cadmium sulfide if 0.7913 g is collected?

Answer 1

CdSO4(aq) + K2S <------> CdS(s) + K2SO4(aq)

Stoichimetrically, 1mole of CdSO4 react with 1mole of K2S

Number of moles = mass/molar mass

given moles of CdSO4 = 1.357g/208.47g/mol = 0.006509mol

given moles of K2S = 0.7519g/110.26g/mol = 0.006819mol

0.006819moles of K2S require 0.006819moles of CdSO4 but available moles of CdSO4 is 0.006509.So, limiting reactant is CdSO4.

Stoichiometrically, 1mole of CdSO4 should give 1mole of CdS

So, number of moles of CdS should be formed = 0.006509

Mass = Number of moles × molar mass

mass of CdS should be formed = 0.006509mol × 144.47g/mol = 0.9404g

Theoretical yield = 0.9404g

percent yield = (Actual yield /Theoretical yield)×100

percent yield = (0.7913g/0.9404g)×100

Percent yield = 84.15%