Problem: How many grams of PbCl2 are formed when 25.0 mL of 0.614 M KCl react with Pb(NO3)2? 2 KCl (aq) + Pb(NO3)2(aq) →2 KNO3 (aq) + PbCl2(s) •21.3 g•12.8 g•8.54 g•4.27 g•2.13 g

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How many grams of PbCl2 are formed when 25.0 mL of 0.614 M KCl react with Pb(NO3)2

2 KCl (aq) + Pb(NO3)2(aq) →2 KNO3 (aq) + PbCl2(s) 

•21.3 g
•12.8 g
•8.54 g
•4.27 g
•2.13 g

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