agcl molar masA 250.0 g sample of a white solid is known to be a mixture of KNO3, BaCl2, and NaCl. When 100.0 g of this mixture is dissolved in water and allowed to react with excess H2SO4, 67.3 g of a white precipitate is collected. When the remaining 150.0 g of the mixture is dissolved in water and allowed to react with excess AgNO3, 197.6 g of a second precipitate is collected. (a) What are the formulas of the two precipitates? (b) What is the mass of each substance in the original 250 g mixture?

a. BaSO₄ and AgCl.

b. 150.0g of BaCl₂, 50.0g of NaCl and 50.0g of KNO₃

Explanation:

Barium, Ba, from BaCl₂ reacts with the SO₄²⁻ of H₂SO₄ to produce BaSO₄, an insoluble white salt.

The reaction is:

BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl

Also, Chlorides from BaCl₂ (2Cl⁻) and NaCl (1Cl⁻) react with AgNO₃ to produce AgCl, another white insoluble salt, thus:

Cl⁻ + AgNO₃ → AgCl + NO₃⁻

a. Thus, formulas of the two precipitates are: BaSO₄ and AgCl

b. Moles of BaSO₄ in 67.3g (Molar mass BaSO₄: 233.38g/mol) are:

67.3g * (1 mol / 233.38g) = 0.2884 moles of BaSO₄ = moles of BaCl₂ Because 1 mole of BaCl₂ produces 1 mole of BaSO₄

Now, as molar mass of BaCl₂ is 208.23g/mol, the mass of BaCl₂ in the mixture of 100.0g is:

0.2884 moles of BaCl₂ ₓ (208.23g / mol) = 60.0g of BaCl₂ in 100g of the mixture

Moles of the AgCl produced (Molar mass AgCl: 143.32g/mol) are:

197.96g ₓ (1mol / 143.32g) = 1.38 moles of AgCl.

As moles of Cl⁻ that comes from BaCl₂ are 0.2884 moles*2*1.5 (1.5 because the sample is 150.0g not 100.0g as in the initial reaction)

= 0.8652 moles of BaCl₂, that means moles of NaCl are:

1.38mol - 0.8652mol = 0.5148 moles of NaCl (Molar mass 58.44g/mol):

Mass NaCl in 150g =

0.5148mol NaCl * (58.44g/mol) = 30.0g of NaCl in 150.0g

That means, in the 250.0g of sample, the mass of BaCl₂ is:

60.0g BaCl₂ ₓ (250.0g / 100g) = 150.0g of BaCl₂

Mass of NaCl is:

30.0g NaCl ₓ (250.0g / 150g) = 50.0g of NaCl

As the total mass of the mixture is 250.0g, the another 50.0g must come from KNO₃, thus, there are 50.0g of KNO₃.