Final answer:
Solid precipitates will form in 2 of the given reactions: when silver nitrate reacts with hydrochloric acid to form silver chloride, and when copper(II) nitrate reacts with sodium hydroxide to form copper(II) hydroxide.
Step-by-step explanation:
To determine how many of the listed reactions will form solid precipitates, let's analyze each one individually based on the solubility rules of ionic compounds in water.
Cu(NO₃)₂ + 2HCl (aq) → CuCl₂ (?) + 2HNO₃ (?) - Copper(II) chloride is soluble in water; no precipitate forms.
AgNO₃ (aq) + HCl (aq) → AgCl (?) + HNO₃ (?) - Silver chloride is insoluble and will precipitate.
Fe(OH)₃ (s) + 3HCl (aq) → FeCl₃ (?) + 3 H₂O (?) - There is no precipitate because Fe(OH)₃ is already a solid and FeCl₃ is soluble.
Cu(NO₃)₂ (aq) + 2 NaOH (aq) → Cu(OH)₂ (?) + 2NaNO₃ (?) - Copper(II) hydroxide is insoluble, hence it will form a precipitate.
Cu(OH)₂ (s) + 4 NH₄OH (aq) → Cu(NH₃)₃4(OH)₂ (?) + 4 H₂O (?) - The complex ion [Cu(NH₃)₃4]^(2+) is soluble in water, so no precipitate will form.
Therefore, solid precipitates will form in 2 of the given reactions: the second and the fourth, making the answer (b) 2.