You could add aquous hydrochloric to the solution to precipitate out silver chloride. What volume of a 0.130 M aqueous hydrochloric solution is needed to precipitate the silver ions from 30.0 mL of a 0.225 M silver nitrate solution? You could add solid potassium chloride to the solution to precipitate silver chloride. What mass of potassium chloride is needed to precipitate the silver ions from 30.0 mL of a 0.225 M silver nitrate solution?

Question

Chemistry

Roderick Graham

29 January, 14:56

You could add aquous hydrochloric to the solution to precipitate out silver chloride. What volume of a 0.130 M aqueous hydrochloric solution is needed to precipitate the silver ions from 30.0 mL of a 0.225 M silver nitrate solution? You could add solid potassium chloride to the solution to precipitate silver chloride. What mass of potassium chloride is needed to precipitate the silver ions from 30.0 mL of a 0.225 M silver nitrate solution?

+1

Answers (1)

Know the Answer?

Manuel Brandt · Answer 1

1) Chemical reaction: AgNO₃ + HCl → AgCl + HNO₃.

V (AgNO₃) = 30,0 mL = 0,03 L.

c (AgNO₃) = 0,225 mol/L.

n (AgNO₃) = 0,03 L · 0,225 mol/L.

n (AgNO₃) = 0,00675 mol.

From chemical reaction: n (AgNO₃) : n (HCl) = 1 : 1.

0,00675 mol : n (HCl) = 1 : 1.

n (HCl) = 0,00675 mol.

V (HCl) = n (HCl) : c (HCl).

V (HCl) = 0,00675 mol : 0,130 mol/L.

V (HCl) = 0,0519 L = 51,92 ml.

2) 1) Chemical reaction: AgNO₃ + KCl → AgCl + KNO₃.

V (AgNO₃) = 30,0 mL = 0,03 L.

c (AgNO₃) = 0,225 mol/L.

n (AgNO₃) = 0,03 L · 0,225 mol/L.

n (AgNO₃) = 0,00675 mol.

From chemical reaction: n (AgNO₃) : n (KCl) = 1 : 1.

0,00675 mol : n (KCl) = 1 : 1.

n (KCl) = 0,00675 mol.

m (KCl) = n (KCl) · M (KCl).

m (KCl) = 0,00675 mol · 74,55 g/mol.

m (KCl) = 0,503 g.

n - amount of substance.

M - molar mass.