Question

A mixture of BaCl2 and NaCl is analyzed by precipitating all the barium as BaSO4. After addition of an excess of Na2SO4 to a 3.988-g sample of the mixture, the mass of precipitate collected is 2.113 g. What is the mass percentage of barium chloride in the mixture?

Answer 1

Answer:

The mass % of BaCl2 is 47.24 %

Explanation:

Step 1: Data given

Molar mass BaSO4 = 233.38 g/mol

Molar mass BaCl2 = 208.23 g/mol

Step 2: The balanced equation

BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq)

Step 3: Calculate moles BaSO4

Moles BaSO4 = mass BaSO4 / molar mass BaSO4

Moles BaSO4 = 2.113 grams / 233.38 g/mol

Moles BaSO4 = 0.00905 moles

Step 4: Calculate moles BaCl2

For 1 mol BaCl2 we need 1 mol Na2SO4 to produce 1 mol BaSO4 and 2 moles NaCl

For 0.00905 moles BaSO4 we need 0.00905 moles BaCl2

Step 5: Calculate mass BaCl2

Mass BaCl2 = moles BaCl2 * molar mass BaCl2

Mass BaCl2 = 0.00905 moles *208.23 g/mol

Mass BaCl2 = 1.884 grams

Step 6: Calculate % BaCl2

% BaCl2 = (1.884 / 3.988 ) * 100%

% BaCl2 = 47.24 %

The mass % of BaCl2 is 47.24 %