Question

In a single displacement reaction, 38.1 grams of aluminum metal reacts with 72.4 grams of nickel(II) chloride dissolved in 1L of water. The reaction yields 29.8 grams of isolated product. For this reaction, provide the a.) balanced molecular equation (including the phase states), b.) oxidation and reduction half-reactions, c.) net ionic equation, d.) identify the limiting and excess reactants, and e.) percent yield of the product in grams. (Note: the molar mass of nickel(II) chloride is 129.59 g/mol)

Answer 1

a) The balanced molecular equation is

2 Al_(S) + 3 NiCl_{2 (aq)} 2 AlCl_{3 (aq)} + 3 Ni _(S)

b) Oxidation half reaction Al _(S) Al⁺³_(aq) + 3 e^-

Reduction half reaction Ni⁺² _(aq) + 2 e^- Ni _(S)

c) Net ionic equation is 2 Al _(S) + 3 Ni⁺² _(aq) 2 Al⁺³_(aq) + 3 Ni _(S)

d) The balanced molecular equation is

2 Al_(S) + 3 NiCl_{2 (aq)} 2 AlCl_{3 (aq)} + 3 Ni _(S)

2 moles of Al reacts with 3 moles of NiCl₂

2 x 26.98 g of Al reacts with 3 x 129.59 g of NiCl₂

53.96 g of Al reacts with 388.77 g of NiCl₂

Then 38.1 g of Al reacts with (38.1 / 53.96) x 388.77 = 274.50 g of NiCl₂

But we have only 72.4 g of NiCl₂ .

So the limiting reagent is NiCl₂ and the excess reagent is Aluminium metal.

e) The balanced molecular equation is

2 Al_(S) + 3 NiCl_{2 (aq)} 2 AlCl_{3 (aq)} + 3 Ni _(S)

3 moles of NiCl₂ gives with 3 moles of Ni

3 x 388.77 g of NiCl₂ gives 3 x 58.69 g of Ni

1166.31 g of NiCl₂ gives 176.07 g of Ni

72.4 g of NiCl₂ gives (72.4 /1166.31) x 176.07 =72.4 g of Ni

But we got 29.8 g of Ni only.

So percent yield =[ (72.4 - 29.8) x 100 ]/ 72.4 = 58.84 %