I know the final Balanced equation is 2Co(NO3)2 + 6NH3 + 2(NH4)2CO3 + H2O2 → 2[Co(NH3)4CO3]NO3 + 2NH4NO3 + 2H2O . Which is all I need for the Post Lab, but I want to undertstand how to get there in case I have something similar on the exam. What are the step by step procedures for balancing? Co(NO3)2 + NH3 (aq) + (NH4)2CO3 + H2O2 → [Co(NH3)4CO3]NO3 + NH4NO3 + 2 H2O Where the first step might be Co(OH2)62+ + 4 NH3 + CO32- → Co(NH3)4CO3 + 6 H2O This Co(II) complex could then be oxidized by the transfer of an electron to H2O2 to givethe relatively non-reactive Co(III) ion, [Co(NH3)4CO3]+. The preparation of [Co(NH3)5Cl]2+ isaccomplished from the carbonato complex according to the following series of equations:[Co(NH3)4CO3]+ + 2 HCl → [Co(NH3)4(OH2)Cl]2+ + CO2 (g) + Cl-[Co(NH3)4(OH2)Cl]2+ + NH3 (aq) → [Co(NH3)5(OH2)]3+ + Cl-[Co(NH3)5(OH2)]3+ + 3HCl → [Co(NH3)5Cl]Cl2 (s) + H2O + 3H+
I know the final Balanced equation is 2Co(NO3)2 + 6NH3 + 2(NH4)2CO3 + H2O2 → 2[Co(NH3)4CO3]NO3 + 2NH4NO3 + 2H2O . Which is all I need for the Post Lab, but I want to undertstand how to get there in case I have something similar on the exam.
What are the step by step procedures for balancing?
Co(NO3)2 + NH3 (aq) + (NH4)2CO3 + H2O2 → [Co(NH3)4CO3]NO3 + NH4NO3 + 2 H2O
Where the first step might be
- Co(OH2)62+ + 4 NH3 + CO32- → Co(NH3)4CO3 + 6 H2O
This Co(II) complex could then be oxidized by the transfer of an electron to H2O2 to give
the relatively non-reactive Co(III) ion, [Co(NH3)4CO3]+. The preparation of [Co(NH3)5Cl]2+ is
accomplished from the carbonato complex according to the following series of equations:
[Co(NH3)4CO3]+ + 2 HCl → [Co(NH3)4(OH2)Cl]2+ + CO2 (g) + Cl-
[Co(NH3)4(OH2)Cl]2+ + NH3 (aq) → [Co(NH3)5(OH2)]3+ + Cl-
[Co(NH3)5(OH2)]3+ + 3HCl → [Co(NH3)5Cl]Cl2 (s) + H2O + 3H+
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