lead (IV) oxide and hydrochloric acid yield lead (IV) chloride and chlorine gas and water.
The skeleton equation is...
Pb2O5 + HCl ==> PbCl5 + Cl2 + H2O
Can this even be balanced? Have I made a mistake in the skeleton equation?
2007-03-23 17:07:20 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
Thank you so much! That was a very lame of me. I can't believe the roman numeral was messing me up.
2007-03-23 17:37:07 · update #1
well, firstly the roman numeral IV is 4, not 5 (V), so the skeleton equation should be
PbO2 + HCl --> PbCl2 + Cl2 + H20
2007-03-23 17:16:18 · answer #1 · answered by Belle 3 · 1⤊ 1⤋
Pb2O + 2 HCl = 2 PbCl5 + -4 Cl2 + H2O
2007-03-24 00:18:34 · answer #2 · answered by Anonymous · 0⤊ 1⤋
The formula for lead (IV) oxide is PbO2
The normal reaction is as follows:
PbO2(s) + 4HCl (aq) ----> PbCl2 (s) + Cl2 (g) + 2H20 (l)
The above happens when lead (IV) oxide is warmed with conc HCl.
PbCl4 is obtained under special conditions: if conc HCl is cooled in ice-salt and the oxide stirred in. A complex ion (PbCl6)2- is also obtained.
2007-03-24 00:59:17 · answer #3 · answered by brisko389 3 · 1⤊ 1⤋
Yes you made a mistake, in the products PbCl4 instead of PbCl5...
2Pb2O5 + 20HCl ---> 4PbCl4 + 2Cl2 + 10H2O
2007-03-24 00:16:46 · answer #4 · answered by MoMoChan 3 · 0⤊ 1⤋