Answer:
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 <em>Because 1 mole of BaCl₂ produces 1 mole of BaSO₄</em>
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) = <em>30.0g of NaCl in 150.0g</em>
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That means, in the 250.0g of sample, the mass of BaCl₂ is:
60.0g BaCl₂ ₓ (250.0g / 100g) = <em>150.0g of BaCl₂</em>
Mass of NaCl is:
30.0g NaCl ₓ (250.0g / 150g) =<em> 50.0g of NaCl</em>
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As the total mass of the mixture is 250.0g, the another 50.0g must come from KNO₃, thus, there are <em>50.0g of KNO₃.</em>
