no of moles of ZnCl2Â Â = W/G.M.Wt
                              Â
= 39.4/136.286Â Â = 0.289moles
no of moles of K2CO3 = molarity * volume in L
                                Â
= 0.6*0.25Â Â Â = 0.15moles
ZnCl2(aq) + K2CO3(aq) ---------> ZnCO3(s) + 2KCl(aq)
1 moles of K2CO3 react with 1 mole of ZnCl2
0.15moles of K2CO3 react with 0.15 moles of ZnCl2 is
required
ZnCl2 is excess reactant
The no of moles of excess reactant left after complete the
reaction = 0.289-0.15Â Â = 0.139moles
molarity of ZnCl2Â Â = no of moles/volume in L
                        Â
= 0.139/0.25Â Â = 0.56M
ZnCl2(aq) -----------> Zn^2+(aq) + 2Cl^- (aq)
0.56M ------------------ 0.56M
The final molarity of Zn^2+ = 0.56M >>>>answer
