Question

CoCl2.6H2O + H2C2O4.2H2O -> CoC2O4.4H2O + 2 HCl + 4 H2O

If, the reaction uses (1.78x10^0) g of CoCl2.6H2O and 1.26 of H2C2O4.2H2O, what is the theoretical yield of CoC2O4.4H2O? This time you have to determine the limiting reagent before determining the theoretical yield.

Answer 1

Molar mass of CoCl₂.6H₂O = 237.93 g/mol

Mass of CoCl₂.6H₂O = 1.78 g

Number of moles of CoCl₂.6H₂O = Mass/molar mass = 1.78/237.93 g/mol = 0.007481 moles

Molar mass of H₂C₂O₄.2H2O = 126.07 g/mol

Mass of H₂C₂O₄.2H2O = 1.26 g

Number of moles of H₂C₂O₄.2H2O = 1.26/126.07 = 0.009994 moles

Since, moles of CoCl₂.6H₂O are in less amount, so, it is limiting reactant

Molar mass of CoC₂O₄.4H₂O = 219.01 g/mol

Theoretical yield = (Molar mass of CoC₂O₄.4H₂O/Molar mass of CoCl₂.6H₂O) x Mass of CoCl₂.6H₂O

= (219.01/237.93) x 1.78

= 1.64 g