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?

The mass percentage of barium chloride in the mixture 47.72 %

The balanced equation is as follows:

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

molar mass of BaCl₂ = 208.23 g /mol

molar mass of  BaSO₄ = 233.38 g/mol

Na₂SO₄ is in excess in the reaction.

Therefore, BaCl₂ will determine the amount of product produced.

233.38 of BaSO₄ requires 208.23 g of BaCl₂

2.133 g of BaSO₄ will require ? BaCl₂

cross multiply

mass of BaCl₂ = 208.23 × 2.133 / 233.38

mass of BaCl₂ = 444.15459 / 233.38

mass of BaCl₂ = 1.90313904362

mass of BaCl₂ ≈ 1.903 grams

% mass of BaCl₂ in the mixture = 1.903 / 3.988 × 100

% mass of BaCl₂ in the mixture = 190.3 / 3.988

% mass of BaCl₂ in the mixture = 47.7216410135

% mass of BaCl₂ in the mixture = 47.72 %

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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 %

its going to be mutualism cuz they both benefit from the relationship.

~batmans wife dun dun

it means there are 3 atoms of oxygen