Final answer:
To calculate the grams of barium chloride reacting with 188 grams of aluminum sulfate, the molar mass of each compound is used to convert grams to moles, then the stoichiometry from the balanced reaction is applied to find that 342.89 grams of barium chloride will react.
Step-by-step explanation:
To calculate how many grams of barium chloride will react with 188 grams of aluminum sulfate, we must first write out the balanced chemical equation for the double displacement reaction:
3 BaCl₂ (aq) + Al₂(SO₄)₃ (aq) → 3 BaSO₄ (s) + 2 AlCl₃ (aq)
Next, we need to find the molar mass of aluminum sulfate and barium chloride. The molar mass of Al₂(SO₄)₃ is 342.15 g/mol, and the molar mass of BaCl₂ is 208.23 g/mol.
Then convert the mass of aluminum sulfate to moles:
- 188 g Al₂(SO₄)₃ × (1 mol Al₂(SO₄)₃ / 342.15 g Al₂(SO₄)₃) = 0.549 moles Al₂(SO₄)₃
According to the balanced equation, we need 3 moles of BaCl₂ for every mole of Al₂(SO₄)₃:
0.549 moles Al₂(SO₄)₃ × (3 moles BaCl₂ / 1 mole Al₂(SO₄)₃) = 1.647 moles BaCl₂.
Finally, convert moles of BaCl₂ to grams:
- 1.647 moles BaCl₂ × (208.23 g BaCl₂ / 1 mol BaCl₂) = 342.89 grams BaCl₂
Therefore, 342.89 grams of barium chloride will react with 188 grams of aluminum sulfate.