Cobalt iii hydroxide and nitric acid react according to the following balanced equation: co(oh)3 (aq) + 3 hno3 (aq) --> co(no3)3(aq) + 3 h2o(l) a. what mass of cobalt iii hydroxide is needed to make 5.2 liters of a 0.42 m cobalt iii hydroxide solution? b. how many milliliters of 1.6 m nitric acid is needed to completely react with the amount of cobalt iii hydroxide in part 2a above? c. based on parts 2a and 2b above, how many moles of water would be produced?

Q1)
molarity is defined as the number of moles in a volume of 1 L.
the molarity of the solution to be prepared is 0.42 M
The volume of the solution to be prepared is 5.2 L
the number of moles of Co(OH)₃ in 1 L solution - 0.42 mol
therefore number of moles required in 5.2 L - 0.42 mol/L x 5.2 L = 2.184 mol
molar mass of Co(OH)₃ - 110 g/mol
mass of Co(OH)₃ required to prepare the solution - 2.184 mol x 110 g/mol = 240.2 g
mass of Co(OH)₃ required - 240.2 g

Q2)
the balanced equation for the reaction between Co(OH)₃ and HNO₃ is;
Co(OH)₃ + 3HNO₃ --> Co(NO₃)₃ + 3H₂O
stoichiometry of Co(OH)₃ to HNO₃ is 1:3
the number of Co(OH)₃ moles present - 2.184 mol
the number of HNO₃ moles required - 2.184 x 3 = 6.552 mol
molarity of HNO₃ is 1.6 M
If 1.6 mol is in 1 L of solution 
then volume containing 6.552 mol is - 6.552 mol / 1.6 mol/L = 4.095 L
therefore volume of HNO₃required is 4095 mL 

Q3)
the balanced equation
Co(OH)₃ + 3HNO₃ --> Co(NO₃)₃ + 3H₂O
stoichiometry of Co(OH)₃ to H₂O is 1:3
the number of Co(OH)₃ moles reacted - 2.184 mol 
when 1 mol of Co(OH)₃ reacts - 3 mol of H₂O is formed
therefore when 2.184 mol of Co(OH)₃ moles react - 2.184 x 3 = 6.552 mol of H₂O formed
therefore number of H₂O moles formed - 6.552 mol

The problem here is that according to aufbau principle before 3d orbital 4s subshell comes.after filling 3p orbital with electrons you have 4 electrons left, you place two in 4s with opposite spin and the other two in 3d in any of its two orbitals of total five orbitals with the same spin.