Precipitate Q?
chemical equation
1CuCl2(aq) + 1Pb(NO3)2(aq) ---- 1Cu(NO3)2(aq) + 1PbCl2(s)
overall ionic equation
1Cu2+(aq) + 2Cl-(aq) + 1Pb2+(aq) + 2NO3-(aq) ----- 1Cu2+(aq) + 2NO3-(aq) + 1PbCl2(s)
net ionic equation
1Pb2+(aq) + Cl-(aq) 1PbCl2(s)
If 19.81 g of copper(II) chloride react, what is the maximum amount of precipitate that could be formed?
Answer:
Molar mass CuCl2 = 134.4 g/mol
19.81 g / 134.4 = 0.147 mol
the ratio between CuCl2 and PbCl2 is 1 : 1 so we get 0.147 mol PbCl2 ( mm = 278.1 g/mol )
0.147 x 278.1 = 40.9 g
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