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

The balanced reaction is 2A (s)l+3NiCl2 (aq) --------->2AlCl3 (aq)+ 3Ni (s) (1)

2Al + 3NI+2 + 6Cl- ----------> 2Al+3+6Cl- + 3Ni ionic equation

3Ni+2+ 6Cl- ------------> 2Al+3+ 3Ni(s) net ionic equation

Al --------->Al+3+3e- oxidation and

Ni+2 +2e- ---------->Ni reduction

The reaction -1 suggests that 2 mole of al reacts with 3 moles of NiCl2 to gives Ni(s), isolated product

Theoretical moles ratio of Al : NiCl2 = 2:3 = 2/3 :1 or 0.66 :1

moles of Al given = mass of Al/atomic weight= 38.1/27=1.41

moles of NiCl2= mass of NiCl2/ molar mass of NiCl2= 72.4/129.59 =0.56

actual moles ratio of Al : NiCl2 = 1.41 :0.56 or 1.41/0.56 : 1 =2.52 :1

so excess is Al as it s required at 0.66 per moles of NiCl2 where as supplied is 2.52. Hence NiCl2 is limiting reactant.

3 moles of NiCl2 correspond to 3*129.59 gm of NiCl2 ( moles= mass/molar mass, mass= moles* molar mass) =388.77 gm of NiCl2

3 mole of NICl2 or 388.77 gm of NiCl2 gives 3 moles of Ni ( atomic weight of NI= 58.59 g/mole)

388.77 gm of NiCl2 gives 3*58.59 = 175.77 gm of Ni

72.4 gm of NiCl2 gives 72.4*175.77/388.77=32.73 gm of Ni, which is the theoretical yield of an isolated product since Ni is solid and gets precipiated.

% yield =100* actual yield/ theoretical yield=100*29.8/32.73=91.04%