Pb(NO3)2 (aq) + 2 KCl (aq) PbCl2 (s) + 2 KNO3 (aq)
If 54.5mL of 3.82M Pb(NO3)2 reacts with 75.3mL of 5.89M KCl react:
Find the limiting reactant.
What is the mass of the precipitate that should be made?
If this reaction has an efficiency of 78.4% what mass of the precipitate would typically be made?
Pb(NO3)2 (aq) + 2 KCl (aq) PbCl2 (s) + 2 KNO3 (aq) If 54.5mL of 3.82M...
How many grams of PbCl2 are formed when 35.0 mL of 0.520 M KCl react with Pb(NO3)2? 2KCl(aq) + Pb(NO3)2(aq) → 2KNO3(aq) + PbCl2(s) How many grams of PbCl2 are formed when 35.0 mL of 0.520 M KCl react with Pb(NO3)2? 2KCl(aq) + Pb(NO3)2(aq) → 2KNO3(aq) + PbCl2(s) 10.1 g 25.3 g 5.06 g 15.2 g 2.53 g
Lead ions can be precipitated from solution with KCl according to the following reaction: Pb2+(aq)+2KCl(aq) →PbCl2(s)+2K+(aq) When 28.5 g KCl is added to a solution containing 25.5 g Pb2+, a PbCl2 precipitate forms. The precipitate is filtered and dried and found to have a mass of 29.3 g. Determine the limiting reactant. Determine the theoretical yield of PbCl2. Determine the percent yield for the reaction.
2NaCl(aq) + Pb(NO3)2(aq)-NaNO3(aq) + PbCl2(s) How many formula units of Pb(NO3)2 would be needed in the above reaction to produce 555kg of PbCl2?
Answer the following questions based on this reaction: Pb(NO3)2 (aq) + 2 NaCl (aq) PbCl2 (s) + 2 NaNO3 (aq) a) If 225 mL of 12.95 M Pb(NO3)2(aq) are reacted with a solution made with 5.05 g of NaCl (aq), how many grams of lead (II) chloride will be precipitated? b) If the actual yield of lead (II) chloride is 1.06 g, what is the percent yield?
Lead ions can be precipitated from solution with NaCl according to the reaction: Pb(NO3)2 (aq) + 2NaCl(aq) → PbCl2(s) + 2NaNO3 (aq) Determine the limiting reactant for the reaction between Pb(NO3)2 and NaCl when 0.155L of 1.5M Pb(NO3)2 is mixed with 0.075L of 1.4M NaCl.
What mass of PbCl2 can form from 0.235L of 0.110 M KCl solution if there is plenty of Pb(NO3)2 present? Molar mass of PbCl2 is 278.1g/mol Balanced reaction: 2KCl(aq) + Pb(NO3)2(aq) -> PbCl2(s) + 2KNO3(aq)
Formula of precipitate Pb(NO3)2(aq) + Kl(aq) → NaOH(aq) + KNO3(aq) → NaOH(aq) + Fe(NO3)2(aq) →
Fe(NO3)3 reacts with K2CrO4 to produce Fe2(CrO4)3 according to the equation 2 Fe(NO3)3 (aq) + 3 K2CrO4 (aq) → Fe2(CrO4)3 (s) + 6 KNO3 (aq) ( a) Which reactant is limiting if 3.744 g of Fe(NO3)3 and 3.825 g of K2CrO4 are allowed to react? (b) What mass of Fe2(CrO4)3 can be produced? (c) What mass of the excess reactant remains when the reaction is complete?
What volume of 1.06 M KCl solution, in mL, is necessary to completely react with 22.98 mL of 0.243 M Pb(NO3)2? 2 KCl(aq) + Pb(NO3)2(aq) → PbCl2(s) + 2 KNO3(aq)
Pb(NO3)2 (0) + 2 KCl(aq) PbCl(s) + 2 KNO₃(aq) • How many mL of a 2.00M Pb (NO3)2 Solution will react tw 50.0mL of a 1.som kce solution?