When What is the molecular formula of the precipitate? AgNO3 23.54 mL of 0.342 M

Question

When What is the molecular formula of the precipitate? AgNO3 23.54 mL of 0.342 M silver nitrate is combined with 18.61 mL of 0.867 M potassium chloride, a precipitate forms. Submit Answer Incorrect. Tries 1/3 Previous Tries What is the molecular formula of the limiting reactant? AgNO3 You are correct. Your receipt no. is 167-4915 Previous Tries What mass of precipitate, in g, is formed? 1.15 You are correct. Your recelpt no. is 167-734Previous Tries What is the molecular formula of the excess reactant? KCI You are correct. r receipt no. is 167-4706Previous Tries What is the concentration, in M, of K+ in the solution after the reaction? .191 Submit Answer Incorrect. Tries 1/3 Previous Tries What is the concentration, In M, of Cl in the solution after the reaction? 0.192 ou are correct. Your receipt no. is 167-5802 What is the concentration, in M, of Ag in the solution after the reaction? -383 Submit Answer what is the concentration, in M, of NO3. in the solution after the reaction?'-383 Previous Ties Incorrect. Tries 2/3 Previous Tries Submit Answer Incorrect. Tries 2/3 Previous Tries

Explanation / Answer

balance equation

AgNO3   + KCl -------> AgCl + KNO3

1mol 1mol 1mol 1mol

First we need to calculate the mols of AgNO3 and KCl

mol of AgNO3 = molarity × Volume = 0.342 mol/L × 0.02354 L = 0.00805 mol

mol of KCl = molarity × Volume = 0.867 mol/L × 0.01861 L = 0.01613 mol

> The molecular formula of precipitate is AgCl

> The molecular formula of limiting reagent is AgNO3

The mols of AgNO3 are limited and the mols of KCl are in excess, so, the limiting reagent is AgNO3

> Mass of precipitate is 1.153 g

From balance equation we know that 1 mol of AgNO3 will reacts with 1 mol of KCl to yields 1 mol of AgCl. So, 0.00805 mol of AgNO3 will reacts will reacts with 0.00805 mol of KCl and yields 0.00805 mol of AgCl.

hence

Mass of precipitate = mol of AgCl × Molar mass of AgCl = 0.00805 mol × 143.32 g/mol = 1.153 g

> The molecular formula of excess reagent is KCl

> The concentration of K+ after reaction

mol of K+ after reaction = mol before the reaction. So, M = mol/ total volume = 0.01613/0.04215 = 0.383 M

> The concentration of Cl-

mol of Cl- after reaction = mol before the reaction - mol of Cl- consumed = 0.01613 - 0.00805 = 0.00808 mol. So,

M = mol of Cl- after reaction/ total volume = 0.00808 mol/0.04215 = 0.192 M

> The concentration of Ag+ = 0 [All the Ag+ are consumed]

> The concentration of NO3- after reaction

mol of NO3- after reaction = mol before the reaction. So, M = mol/ total volume = 0.00805/0.04215 = 0.191 M

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.