Respuesta :
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 %
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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