Question

Write the balanced net ionic equation.

1.)solid Fe(OH)3and excess 3.0 M HC2H3O2

2.)0.10 M NaHCO3and excess 0.050 M H2SO4

Answer 1

1- The chemical reaction between Fe(OH)_3(s) and HC₂H₃O_2(aq) is

Fe(OH)_3(s) + 3 HC₂H₃O_2(aq) ----------------> Fe(C₂H₃O₂)_3(aq) + 3H₂O_(l)  

Now here the Fe(C₂H₃O₂)_3(aq) formed is a water soluble salt. Again in the reactant, the acid HC₂H₃O_2(aq) is also water soluble. SO they dissociate to their constituent ions . So

Ionic equation is

Fe(OH)_3(s) + 3 H⁺_(aq) + 3C₂H₃O₂^-_(aq) ----------------> Fe⁺³_(aq) + 3(C₂H₃O₂)_3(aq) + 3H₂O_(l)  

Now net ionic equation is formed by cancelling the common ions from the reactant and product side.i.e

Net ionic equation is-

Fe(OH)_3(s) + 3 H⁺_(aq) ----------------> Fe⁺³_(aq) + 3H₂O_(l)  

2- Similarly

The chemical reaction between NaHCO_3(aq) and H₂SO_4(aq) is

2 NaHCO_3(aq) + H₂SO_4(aq) ----------------> Na₂SO_4(aq) + 2H₂O_(l) + CO_2(g)

Now here all the reactants and aqueous products are water soluble. SO they dissociate to their constituent ions . So

Ionic equation is

2Na⁺_(aq) + HCO₃^-_(aq) + 2H⁺_(aq) + SO₄^2-_(aq) ----------------> 2Na⁺_(aq) + SO₄^2-_(aq) + 2H₂O_(l) + CO_2(g)

Now net ionic equation is formed by cancelling the common ions from the reactant and product side.i.e

Net ionic equation is-

HCO₃^-_(aq) + 2H⁺_(aq) ----------------> 2H₂O_(l) + CO_2(g)