How many moles of NaCl, if mixed with excess Pb2+ ions in solution, would be needed to form 11.7 g of PbCl2?

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Answer 1

Answer #1

mass of PbCl2 = 11.7 g

moles of PbCl2 = 11.7 / 278.106

= 0.04207 mol

2 NaCl (aq) + Pb2+ (aq) --------------> PbCl2 (s) + 2 Na+

2 1 1

?? 0.04207 mol

moles of NaCl = 0.04207 x 2

moles of NaCl = 0.0841 mol

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Answer 2

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