2006-06-24 15:55:49 · 4 answers · asked by lune_ellise 3 in Science & Mathematics ➔ Chemistry
You have two ionic substances, it will undergo a double displacement reaction.
Pb(C2H3O2)2 + 2 NaCl --> PbCl2 + 2 NaC2H3O2
Now since Lead(II) acetate, Sodium chloride, and sodium acetate are soluble, they break up in water, and the net ionic equation is
Pb2+ + 2 Cl- --> PbCl2
PbCl2 or Lead (II) chloride, is the precipitate formed, and it's generally yellowish.
2006-06-24 22:37:37 · answer #1 · answered by beekay36 2 · 1⤊ 0⤋
Pb(CH3COO)2 + 2NaCl === 2NaCH3COO + PbCl2
sodium acetate lead chloride
2006-06-24 16:00:35 · answer #2 · answered by Anonymous · 0⤊ 0⤋
lead (II) chloride and sodium acetate
2006-06-24 16:00:02 · answer #3 · answered by dkrgrand 6 · 0⤊ 0⤋
2NaCl (aq) + Pb(CH3COO)2 (aq) ---> Pb(II)Cl2 (s) + 2 Na+ (aq)+ 2 CH3COO- (aq) This is a double displacement reaction and Na ions as well as the acetate ions are soluble
2016-03-27 03:28:53 · answer #4 · answered by Anonymous · 0⤊ 0⤋