Answer to Question #161942 in Chemistry for Egwolor Naomi

Question

0.2025g sample consisting of only barium two chloride and potassium chloride required 20.25ml of 0.1200mole of AgNO3 solutionforthe quantitative precipitationof chloride. Calculate the percentage of barium and percentage of calcium

Answer 1

The molar mass of BaCl₂ is 208.23 g mol^-1;

The molar mass of KCl is 74.55 g mol^-1;

The molar mass of Ba is 137.33 g mol^-1;

The molar mass of K is 39.098 g mol^-1.

Solution:

let x = mass of KCl, y = mass of BaCl₂

Thus: x + y = 0.2025 g

Or: x = 0.2025 - y

Moles of AgNO₃ = Molarity of AgNO₃ × Solution volume

Moles of AgNO₃ = 0.1200 M × 0.02025 L = 0.00243 mol

Moles of AgNO₃ = 0.00243 mol

Balanced chemical equations:

KCl + AgNO₃ → AgCl + KNO₃

BaCl₂ + 2AgNO₃ → 2AgCl + Ba(NO₃)₂

According to these equations:

Moles AgNO₃ from KCl = (x g KCl × 1 mol AgNO₃) / (74.55 g mol^-1 KCl) = 0.01341x mol AgNO₃

Moles AgNO₃ from BaCl₂ = (y g BaCl₂ × 2 mol AgNO₃) / (208.23 g mol^-1 BaCl₂l) = 0.00960y mol AgNO₃

0.01341x mol AgNO₃ + 0.00960y mol AgNO₃ = 0.00243 mol AgNO₃

0.01341x + 0.00960y = 0.00243

0.01341 × (0.2025 - y) + 0.00960y = 0.00243

0.002716 - 0.01341y + 0.00960y = 0.00243

-0.0038y = -0.00027

y = 0.07506 g - mass of BaCl₂

x = 0.2025 - y = 0.2025 - 0.07506 = 0.12744

x = 0.12744 g - mass of KCl

%BaCl₂ = (0.07506 g / 0.2025 g) × 100% = 37.07%

%KCl = 100% - %BaCl₂ = 100% - 37.07% = 62.93%

Moles of BaCl₂ = Mass of BaCl₂ / Molar mass of BaCl₂

Moles of of BaCl₂ = 0.07506 g / 208.23 g mol^-1 = 0.0003605 mol BaCl₂

BaCl₂ → Ba²⁺ + 2Cl^-

Hence,

Moles of Ba²⁺ = Moles of BaCl₂ = 0.0003605 mol

Mass of Ba²⁺ = Moles of Ba²⁺ × Molar mass of Ba²⁺

Mass of Ba²⁺ = 0.0003605 mol × 137.33 g mol^-1 = 0.04951 g

%Ba = (0.04951 g / 0.2025 g) × 100% = 24.45%

%Ba = 24.45%

Moles of KCl = Mass of KCl / Molar mass of KCl

Moles of of KCl = 0.12744 g / 74.55 g mol^-1 = 0.001709 mol KCl

KCl → K⁺ + Cl^-

Hence,

Moles of K⁺ = Moles of KCl = 0.001709 mol

Mass of K⁺ = Moles of K⁺ × Molar mass of K⁺

Mass of K⁺ = 0.001709 mo × 39.098 g mol^-1 = 0.06682 g

%K = (0.06682 g / 0.2025 g) × 100% = 32.99% = 33.00%

%K = 33.00%

Answer:

%Ba = 24.45%;

%K = 33.00%.