Write the reaction equations in molecular and ionic form corresponding to the following sequence of transformations

Write the reaction equations in molecular and ionic form corresponding to the following sequence of transformations: Co (OH) 2 = CoOHCl = CoCl2 = Co (NO3) 2

Carry out transformations:
Co (OH) 2 → CoOHCl → CoCl2 → Co (NO3) 2
1) Add an insufficient amount of hydrochloric acid to cobalt (II) hydroxide.
Let’s write the reaction equation in molecular form:
Co (OH) 2 + H Cl = CoOHCl + H2O, the exchange reaction proceeds to the end, because water is released.
Let us write the reaction equation in full ionic form:
Co (OH) 2 + H ^ (+) + Cl ^ (-) = Co ^ (2 +) + OHCl ^ (2-) + H2O
2) Add hydrochloric acid to cobalt (II) hydroxychloride – CoOHCl.
Let’s write the reaction equation in molecular form:
CoOHCl + HCl = CoCl2 + H2O, the exchange reaction proceeds to the end, because water is released.
Let us write the reaction equation in full ionic form:
Co ^ (2 +) + OHCl ^ (2-) + H ^ (+) + Cl ^ (-) = Co ^ (2 +) + 2 Cl ^ (-) + H2O
Let us write the reaction equation in the abbreviated ionic form:
OHCl ^ (2-) + H ^ (+) = Cl ^ (-) + H2O
3) Add silver (I) nitrate (AgNO3) to cobalt (II) chloride (CoCl2).
Let’s write the reaction equation in molecular form:
CoCl2 + AgNO3 = AgCl + Co (NO3) 2
Let’s arrange the coefficients:
CoCl2 +2 AgNO3 = 2 AgCl + Co (NO3) 2, the exchange reaction proceeds to the end, because a precipitate of silver (I) chloride (AgCl) is precipitated.
Let’s compose the complete ionic reaction equation:
Co ^ (2 +) + 2 Cl ^ (-) + 2 Ag ^ (+) + 2NO3 ^ (-) = 2 AgCl + Co ^ (2+) + 2NO3 ^ (-)
Let’s compose an abbreviated ionic reaction equation:
2 Cl ^ (-) + 2 Ag ^ (+) = 2 AgCl
Reduce the coefficients by 2 and get
Cl ^ (-) + Ag ^ (+) = AgCl