Precipitate Q?

Copper(II) chloride and lead(II) nitrate react in aqueous solutions by double replacement. Write the balanced chemical equation, the overall ionic equation, and the net ionic equation for this reaction. Use the lowest possible coefficients.

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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