Question

A student is given a mixture of NaCl(s) and NaNO3(s) and is tasked with determining the percent of NaCl in the mixture. The student dissolves 3.613 g of the mixture in 50 mL of DI water. The student then adds excess AgNO3(aq) to precipitate the chloride ion as AgCl(s). The student determines that 2.268 g of AgCl is formed.

Answer 1

Answer:

25.60% of NaCl in the mixture

Explanation:

With the amount of AgCl produced we can now the moles of Cl⁻ = Moles of NaCl. With these moles we can now the mass of NaCl and, as total mass was 3.613g, it is possible to determine mass percent of NaCl in the mixture.

Moles NaCl:

2.268g AgCl * (1mol / 143.32g AgCl) = 0.01582 moles AgCl

Molar mass AgCl 143.32g/mol

0.01582 moles AgCl = Moles Cl⁻ = Moles NaCl

Mass NaCl (Molar mass: 58.44g/mol):

0.01582 moles NaCl * (58.44g / mol) = 0.9248g NaCl in the mixture

Mass percent:

(0.9248g NaCl / 3.613g) * 100 =

25.60% of NaCl in the mixture