You perform a reaction in which 100.0 mL of 0.200 M lithium hydroxide is mixed into 200.0 mL of 0.150 nickel ii sulfate

Question

A)Determine the limiting reactant

B)Determine the theoretical yield of nickel ii hydroxide

C)Determine the actual yield of nickel ii hydroxide given that the filter paper has a mass of 0.98 grams and the filter paper+nickel ii hydroxide has a mass of 1.95 grams

D)Determine the % yield of nickel ii hydroxide

Answer 1

2LiOH + NiSO₄ ==> Li₂SO₄ + Ni(OH)₂(s) (this assumes nickel(II) hydroxide is insoluble; it is partly soluble)

To determine limiting reactant, find moles of each reactant and divide by its coefficient:

moles LiOH = 0.1 L x 0.200 mol/L = 0.0200 moles LiOH (divided by 2 -->0.0100)

moles NiSO4 = 0.2 L x 0.150 mol/L = 0.0300 moles NiSO4 (divided by 1--> 0.0300)

LIMITING reactant is LiOH

Theoretical yield:

0.0200 moles LiOH x 1 mole Ni(OH)2/2 moles LiOH x 92.7 g/mole = 0.927 g Ni(OH)2

Actual yield:

1.95 g - 0.98 g = 0.97 g (NOTE: This is greater than the theoretical yield of 0.927 g. This is not possible, as % yield will then be greater than 100%). If we assume that the nickel(II) hydroxide formed and collected is the monohydrate, then it will have a molar mass of 110.7 and theoretical yield will be

0.0200 mol LiOH x 1 mol Ni(OH)2/2 mol LiOH x 110.7 g/mole = 1.107 g

Percent yield (assuming the monohydrate of nickel(II)hydroxide:

0.97 g/1.107 g (x100%) = 91%