For the reaction: Pb(NO3)2(aq) + 2KCl(aq)->PbCl2(s) + 2KNO3(aq)
How many grams of PbCl2 will be formed from 63.0 mL of a 1.60 M KCl solution?

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  1. mols KCl = M x L = ?
    mols PbCl2 formed = ?mols KCl x (1 mol PbCl2/2 mols KCl) = ?mols KCl x 1/2 = ? mols PbCl2.
    Then grams PbCl2 = mols PbCl2 x molar mass PbCl2.

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