What will be the result is 100mL of 0.06M of Mg(NO3)2 is added in 50mL of 0.06M of Na2C2O4? Assume the reaction is taking place at 25C. Ksp(MgC04) at 25C-8.56x10-5 O No precipitate will form O A precipitate will form and both Mg2+ and CO42- ions are present in excess O A precipitate will form and an excess of Mg2+ ions will remain in solution O A precipitate will form and an excess of C042 ions will remain in solution