Question

In: Chemistry

In a single displacement reaction, 38.1 grams of aluminum metal reacts with 72.4 grams of nickel(II)...

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)

Solutions

Expert Solution

a) The balanced molecular equation is

2 Al(S) + 3 NiCl2 (aq) 2 AlCl3 (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 NiCl2 (aq) 2 AlCl3 (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 NiCl2 (aq) 2 AlCl3 (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 %


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