Problem 5 When 23.54 mL of 0.342 M silver nitrate is combined with 18.61 mL of 0

Question

Problem 5 When 23.54 mL of 0.342 M silver nitrate is combined with 18.61 mL of 0.867 M potassium chloride, a precipitate forms What is the molecular formula of the precipitate? Submit Answer Tries 0/3 What is the molecular formula of the limiting reactant? Submit Answer Tries 0/3 What mass of precipitate, in g, is formed? Submit Answer Tries o/3 What is the molecular formula of the excess reactant? Submit Answer Tries 0/3 What is the concentration, in M, of K in the solution after the reaction? Submit Answer Tries 0/3 What is the concentration, in M, of Cl in the solution after the reaction? Submit Answer Tries 0/3 What is the concentration, in M, of Ag in the solution after the reaction? Submit Answer Tries o/3 What is the concentration, in M, of NOs in the solution after the reaction? Submit Answer Tries 0/3

Explanation / Answer

AgNO3= 23.54 ml of 0.342M

number of moles of AgNO3= 0.342Mx0.02354L= 0.00805 moles

KCl= 18.61 ml of 0.867M

number of moles of KCl= 0.867M x 0.01861L= 0.0161 moles

AgNO3(aq)    + KCl(aq) --------------- AgCl(s)   + KNO3(aq)

number of moles of AgNO3 is lower thanthe numner moles of KCl. so AgNO3 is used up in the reaction first.

Hence AgNO3 is limiting reagent.

a) The molecular formula of the limiting reagent = AgNO3

b) According to equation

1 mole of AgNO3= 1 mole of AgCl

0.00805 mole of AgNO3 = 0.00805 mole of AgCl

number of moles of AgCl= 0.00805 mole

molar mass of AgCl= 143.32 gram/mole

mass of 0.00805 mole of AgCl= 0.00805 x 143.32= 1.1537

mass of Precipitate formed= 1.154 grams.

C) The molecular formula of excess reagent = KCl

d)

Total volume of the solution = 23.54 +18.61 =42.15 ml

after mixing . the concentration of KCl = 23.54 x0.342/42.15 = 0.191 M

concentration of K+ ion = 0.191M

concentration of Cl- ion = 0.191M

after mixing concentration of AgNO3 = 18.61 x 0.867/42.15 = 0.383M

after mixing

concentration of Ag+ = 0.383M

concentration of NO3- = 0.383M

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