Asked by DOnnie
Calculate the mass of PbCl2 formed when an excess of 0.100 M solution of NaCl is added to 0.4L of 0.7 M Pb(NO3)2.
Pb(NO3)2 + 2 NaCl --> PbCl2 + 2 NaNO3
Pb(NO3)2 + 2 NaCl --> PbCl2 + 2 NaNO3
Answers
Answered by
DrBob222
mols Pb(NO3)2 = M x L = approx 0.28
mols PbCl2 formed = mols Pb(NO3)2 (see the 1:1 ratio for Pb(NO3)2/PbCl2
Then g PbCl2 = mols PbCl2 x molar mass PbCl2.
mols PbCl2 formed = mols Pb(NO3)2 (see the 1:1 ratio for Pb(NO3)2/PbCl2
Then g PbCl2 = mols PbCl2 x molar mass PbCl2.
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