i need help for I J and K. 2 CoCl2 + 2 NH4Cl + H2O2 + 10 NH3 ----> 2 Co(NH3)6Cl3

Question

i need help for I J and K.

2 CoCl2 + 2 NH4Cl + H2O2 + 10 NH3 ----> 2 Co(NH3)6Cl3 + 2 H2O

1) You are assigned the following amount of CoCl2.
Volume of CoCl2 assigned and used 4.94 mL
Concentration of CoCl2 1.79 M
To this is added...
Mass of charcoal 0.20 g
Volume of 15.0 M ammonia 5.0 mL
Mass of ammonium chloride 1.42 g
The ammonium chloride was dissolved. The mixture was cooled. To this was added:
Volume of 10% (by weight) hydrogen peroxide 4.9 mL
(density = 1.00 g/mL)
Calculate:
(Base theoretical values on millimoles CoCl2 used)
a) Millimoles of CoCl2 used ___8.8426____________ mmol
b) Millimoles of NH3 theoretically required __________44.213_____ mmol
c) Millimoles of NH3 actually added __________75_____ mmol
d) Millimoles of NH4Cl theoretically required ________8.8426_______ mmol
e) Millimoles of NH4Cl actually added ___________26.546____ mmol
f) Millimoles of H2O2 theoretically required ___________4.4213____ mmol
g) Millimoles of H2O2 actually added __________14.40_____ mmol
h) Which of the above is the limiting reagent? __________cobalt_____
i) Theoretical yield of Co(NH3)6Cl3 _______________ g
After reaction, workup, and recrystallization, the dry product was put into a sample tube, labeled, and turned in.
Mass of the empty sample tube 9.4918 g
Mass of sample tube with Co(NH3)6Cl3 11.7599 g
Calculate:
j) Weight of product turned in _____________ g
k) The per cent yield of complex _____________ %

Explanation / Answer

I- 2 CoCl2 + 2 NH4Cl + H2O2 + 10 NH3 ----> 2 Co(NH3)6Cl3 + 2 H2O

     2mol        2mol         1 mol      10 mol              2mol

mole of CoCl2 = 1.79 X 4.94 = 8.8426mmol

mole of NH3 = 15 x5 = 75mmol

mole of NH4Cl = 1.42/53.491= 26.546mmol

mole of H2O2= 14.4

from reaction equation

2 mol of complex is formed when 2 mol of CoCl2 is reacted. So

8.8426 mmol of Co(NH3)6Cl3 is fomed = 8.8426 x10^-3x267.5 =2.365 g

J) Mass of the empty sample tube 9.4918 g
Mass of sample tube with Co(NH3)6Cl3 11.7599 g

Weight of product turned in 11.7599g-9.4918 g = 2.2618g

K) theoretical yield =2.365g

The per cent yield of complex =( practical yield / theoretical yield ) x 100 =( 2.2618 / 2.365)x100= 95.636%

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