Question

A nickel coordination compound, Ni(NH3)6CL2, can be prepared by a procedure similar to the one you will use to prepare Cu(NH3)4SO4 H20. the reaction can be represented stoichiometrically as NiCl2 6H20 + 6 NH4OH rightarrow Ni (NH3)6Cl2 + 12H20 4.850g of NiCl2 6H20 is treated with 22.0 mL of 15M NH40H solution. 4.230g of product is obtained What is the limiting reagent? What is the ratio (moles NH3/moles Ni2+) in solution? Calculate theoretical yield and percent yield.

Answer 1

NiCl2*6H20 + 6NH4OH -----------> Ni(NH3)6 Cl2 + 12H20

Moles of  NiCl2 6H20 = mass / Molar mass = 4.85 / 238 = 0.020378

Moles of NH4OH = molarity*V in L = 0.33

1 mole of NiCl2*6H20 reacts with 6 moles of NH4OH

0.020378 moles of NiCl2*6H20 reacts with = 0.020378*6 =0.1222 moles of NH4OH

Limiting reagent = NiCl2*6H20 (the reactant that consumed fully )

Moles of product = moles of NiCl2 6H20

Mass of product = moles *Molar mass = 0.020378 * 231.6 = 4.72 gms

Theoritical yield = 4.72 gms

% yield = Actual yield*100 / Theoritical yield = 4.23*100 / 4.72 = 89.61%