A solution of NaCl(aq) is added slowly to a solution of lead nitrate, Pb(NO3)2(aq) Pb ( NO 3 ) 2 ( aq ) , until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 10.47 g PbCl2(s) 10.47 g PbCl 2 ( s ) is obtained from 200.0 mL 200.0 mL of the original solution. Calculate the molarity of the Pb(NO3)2(aq) Pb ( NO 3 ) 2 ( aq ) solution. concentration: M
Balanced equation: Pb(NO3)2(aq) + 2NaCl(aq) ----> PbCl2(s) + 2NaNO3(aq)
The moles of PbCl2 = 10.47 g/(278.1 g.mol-1) = 0.03765 mol
From the balanced equation, the mole ratio Pb(NO3)2:PbCl2 = 1:1
i.e. The molarity of Pb(NO3)2 solution = 0.03765 mol/0.2 L = 0.18824 M
Note: 200 mL = 200/1000 = 0.2 L
A solution of NaCl(aq) is added slowly to a solution of lead nitrate, Pb(NO3)2(aq) Pb ( NO 3 ) 2 ( aq ) , until no furth...
8. A solution of NaCl(aq) is added slowly to a solution of lead nitrate, Pb(NO3)2(aq), until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 12.12 g PbCl2(s) is obtained from 200.0 mL of the original solution. Calculate the molarity of the Pb(NO3)2(aq) solution. concentration: M
A solution of NaCl(aq) is added slowly to a solution of lead nitrate, Pb(NO3)2(aq), until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 15.59 g PbCl2(s) is obtained from 200.0 mL of the original solution. Calculate the molarity of the Pb(NO3)2(aq) solution. concentration: M
Suppose we have a solution of lead nitrate, Pb(NO3)2(aq). A solution of NaCl(aq) is added slowly until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 15.93 g of PbCl2(s) is obtained from 200.0 mL of the original solution Calculate the molarity of the Pb(NO3)2(aq) solution. Number
A solution of NaCl(aq) is added slowly to a solution of lead nitrate, Pb(NO), (aq), until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 14.12 g PbCI,(s) is obtained from 200.0 mL of the original solution Calculate the molarity of the Pb(NO), (aq) solution. concentration:
By titration, it is found that 91.9 mL of 0.150 M NaOH(aq) is needed to neutralize 25,0 mL of HCl(aq). Calculate the concentration of the HCl solution. HCl concentration: A solution of NaCl(aq) is added slowly to a solution of lead nitrate, Pb(NO3)2(aq), until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 19.42 g PbCl(s) is obtained from 200.0 mL of the original solution. Calculate the molarity of the Pb(NOx),(aq) solution. concentration:
Question 25 (4 points) Solutions of Pb(NO3)2 and NaCl are mixed to form a solution with final concentrations 0.01 M of Pb(NO3)2 and 0.025 M of NaCl. What will happen once these solutions are mixed? For PbCl, Ksp = 1.7 x 10-5. Sodium nitrate will precipitate, leaving an unsaturated solution of PbCl2 Nothing will happen, there will be no precipitate. Lead chloride will precipitate out of solution, leaving an unsaturated solution of PbCl2 Lead chloride will precipitate out of solution,...
Potassium iodide reacts with lead (ii) nitrate in the following precipitation reaction: 2KI (aq) + Pb(NO3)2 (aq)---> 2KNO3 (aq) + PbI2 (s) What minimum volume of 0.200 M potassium iodide solution is required to completely precipitate all the lead in 155.0 mL of a 0.122 M lead (ii) nitrate solution?
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.
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?
When a solution of ammonium chloride, NH4Cl, is added to a solution of lead (II) nitrate , Pb(NO3)2, a white precipitate, PbCl2, forms. Which of the following is the total ionic equation for this reaction? A) 2 NH4Cl(aq)+Pb(NO3)2(aq)-->PbCl2(s)+2 NH4NO3 (aq) B) 2 NH4+(aq)+2Cl-(aq)+Pb2+(aq)+2NO3-(aq)-->PbCl2(s)+2NH4+(aq)+2NO3-(aq) C) Pb2+(aq)+2Cl-(aq)-->PbCl2(s) D) Pb2+(aq)+Cl2-(aq)-->PbCl2(s)