Zn(OH)2 (zinc hydroxide) is formed
The products in the reaction of Zn(NO3)2 with 2 NaOH are Zn(OH)2(s) and 2 NaNO3 . There are no 'elements' involved
Zn is a semi metal and reacts with acid as well as with alkali. Zn + 2NaOH ---------> Na2ZnO2 + H2
intially: Zn2+ + 2OH- -->Zn(OH)2 excess hydroxide: Zn(OH)2 + 2OH- --> [Zn(OH)4]2- It is true that excess hydroxide reacts further, but the reaction is as under: Zn(OH)2 +2(OH)- ----> (ZnO2) 2- +2H2O or ZnSO4 + 4NaOH ------> 2Na2ZnO2 +2H2O, the white precipitate of zinc hydroxide dissolves with excess hydroxide and forms a soluble complex called sodium zincate. The solution becomes clear. No ppt.
NaOH reacts in a glass wear, forming silicates that might affect the result of the experiment.
ZnCl2 + NaOH --> NaCl2 + ZnOH Zn2+ + Cl- + Na+ +OH- --> Na+ + Cl- + ZnOH Zn+ + OH- ---> ZnOH(s)
The products in the reaction of Zn(NO3)2 with 2 NaOH are Zn(OH)2(s) and 2 NaNO3 . There are no 'elements' involved
There's two possible reactions Zink with solid NaOH gives double salt Zn + 2NaOH --t--> Na2ZnO2 + H2 And I Think you need the other one: Zink with NaOH(aq) gives the complex salt: Zn + 2NaOH + 2H2O --> Na2[Zn(OH)4] + H2
Zn is a semi metal and reacts with acid as well as with alkali. Zn + 2NaOH ---------> Na2ZnO2 + H2
Aluminium hydroxide = Al(OH)3 , and Zinc hydroxide = Zn(OH)2 , is redissolved in excess Sodium hydroxide = NaOH (in water solution) : Al(OH)3 + NaOH + H2O ----> NaAl(OH)4 Zn(OH)2 + NaOH + H2O ----> Na2Zn(OH)4
Simple methods:- reaction of Zn with HCl- reaction of Al with NaOH
intially: Zn2+ + 2OH- -->Zn(OH)2 excess hydroxide: Zn(OH)2 + 2OH- --> [Zn(OH)4]2- It is true that excess hydroxide reacts further, but the reaction is as under: Zn(OH)2 +2(OH)- ----> (ZnO2) 2- +2H2O or ZnSO4 + 4NaOH ------> 2Na2ZnO2 +2H2O, the white precipitate of zinc hydroxide dissolves with excess hydroxide and forms a soluble complex called sodium zincate. The solution becomes clear. No ppt.
2NaCl + Zn(OH)2
Zn(s) + 2 OH-(aq) -> ZnO22-(aq) + H2 (g) Perhaps a bit more may be needed Zn(s) + H2O(l) +2OH-(aq) -> Zn(OH)42-(aq) + H2(g)
NaOH reacts in a glass wear, forming silicates that might affect the result of the experiment.
ZnCl2 + NaOH --> NaCl2 + ZnOH Zn2+ + Cl- + Na+ +OH- --> Na+ + Cl- + ZnOH Zn+ + OH- ---> ZnOH(s)
zinc nitrate + sodium hydroxide yields sodium nitrate and zinc hydroxide( white precipitate)
Please name this analysis for a possible answer.