Problem 5 When 23.54 mL of 0.342 M silver nitrate is combined with 18.61 mL of 0
ID: 712476 • Letter: P
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 0/3Explanation / 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