Precipitate Formation in Reactions

Steps:

• Excess Ca(s) + H₂O(l) → Ca(OH)₂(s) (white precipitate) + H₂(g); produces white solid.
• BaCl₂(aq) + Sr(OH)₂(aq) → Ba(OH)₂(aq) + SrCl₂(aq); all products soluble, no solid.
• CaCO₃(s) + excess HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g); solid dissolves, no precipitate remains.
• MgSO₄(aq) + Ba(NO₃)₂(aq) → Mg(NO₃)₂(aq) + BaSO₄(s) (white precipitate); produces white solid.

Why C is correct:

• Two mixtures (1 and 4) form insoluble white solids per solubility rules: Ca(OH)₂ sparingly soluble, BaSO₄ insoluble.

Why the others are wrong:

• A: Ignores two white precipitates formed.
• B: Understates; exactly two solids produced.
• D: Overstates; only two, not three, yield white solids.

Final answer: C