Question

If your start the experiment with 0.500 grams of Nickel (II) Acetate Tetrahydrate, how many grams of Nickel (II) Formate should you get, assuming 100% yield?

Answer 1

Solution :-

Balanced reaction equation

Ni(CH3COO)2.4H2O + 2 HCOOH ----- > Ni(HCOO)2+ 2CH3COOH + 4 H2O

0.500 g Ni(CH3COO)2.4H2O reacted

how many gram Ni(HCOO)2 form. ?

moles of the Ni(CH3COO)2.4H2O = mass / molar mass

                                                       = 0.500 g / 248.8426 g per mol

                                                       = 0.00201 mol Ni(CH3COO)2.4H2O

now lets calculate the moles of nickel formate using the mole ratio.

0.00201 mol Ni(CH3COO)2.4H2O * 1 mol Ni(HCOO)2 / 1 mol Ni(CH3COO)2.4H2O = 0.00201 mol Ni(HCOO)2

now lets convert the moles of nickecl (ll) formate to its mass

mass= moles* molar mass

       = 0.00201 mol * 148.7283 g per mol

      = 0.399 g Ni(HCOO)2

Threfore by the reaction of the 0.500 g Ni(CH3COO)2.4H2O it can produce 0.299 g nickel (ll) formate.