Chem 122 Esbenshade Sp18 nu Contents Grades Contents ..HW 0> Precipitation calcu

Question

Chem 122 Esbenshade Sp18 nu Contents Grades Contents ..HW 0> Precipitation calculations When 24.83 mL of 0.140 M silver nitrate is combined with 16.58 mL of 0.892 M potassium chloride, a precipitate forms. What is the molecular formula of the precipitate? Submit Answer Tries 0/99 What is the molecular formula of the limiting reactant? Submit Answer Tri es o/99 What mass of precipitate, in g, is formed? Submit Answer Tries 0/99 What is the molecular formula of the excess reactant? Submit Answer Tries 0/99 What is the concentration, in M, of Kt in the solution after the reaction? Submt Answer Tries 0/99 What is the concentration, in M, of Cl in the solution after the reaction? Submit Answer Tries 0/99 What is the concentration, in M, of Ag+ in the solution after the reaction? Submit Answer Tries 0/99 What is the concentration, in M, of NO, in the solution after the reaction? Submit Answer Tries 0/99 Post Discussion 0 Type here to search

Explanation / Answer

silver nitrate,AgNO3

potassium chloride,KCl

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

1)precipitate molecular formula: AgCl

2)mol of AgNO3 reacted=24.83ml*0.140M=0.02483L*0.140mol/L=0.00348 mol

mol of KCl reacted=0.01658L*0.892mol/L=0.0148 mol

they react in 1:1 molar ratio

As AgNO3 is in lesser amount so ,AgNO3 is the limiting reactant

3)mol of precipitate formed=mol of limiting reactant=0.00348mol [since they react in 1:1 molar ratio]

mass of precipitate formed =mol of precipitate formed* molar mass of ppt=0.00348mol*143.32g/mol=0.499=0.5g

4)Molecular formula of excess reactant=KCl

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

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