Which combination will result is a precipitate forming?

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Question restatement: Which combination will result in a precipitate forming?

Option 1: NaOH(aq) and HCl(aq)
This mixture is a neutralization between a strong base (NaOH) and a strong acid (HCl). The products are NaCl and H2O, both highly soluble, with no insoluble solid expected. Therefore, no precipitate should form in typical aqueous conditions.

Option 2: NaOH(aq) and FeCl3(aq)
Mixing a strong base (NaOH) with a soluble iron(III) salt (FeCl3) leads to hydrolysis of Fe3+ and formation of iron(III) hydroxide, Fe(OH)3, which is an insoluble solid that precipitates out of solution. The overall reaction can be represented as FeCl3(aq) + 3NaOH(aq) → Fe(OH)3(s) + 3NaCl(aq). This is a classic precipitation reaction due to the low solubility of Fe(OH)3.

Option 3: Zn(s) and HCl(aq)
This is a redox and acid displacement scenario where zinc metal reacts with hydrochloric acid to produce hydrogen gas and an aqueous zinc chloride. The reaction is Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g). There is no formation of an insoluble solid; thus no precipitate is expected.

Option 4: Na2SO4(aq) and FeCl3(aq)
This mixture contains two soluble salts. In many common aqueous solutions, sodium sulfate and ferric chloride remain in solution, and no insoluble precipitate forms. Although sulfate-containing salts can precipitate under certain conditions with specific cations, Fe2(SO4)3 or other ferric sulfate species are typically soluble, so a precipitate is not anticipated here.

Option 5: None of these
This option would be selected if none of the stated combinations produced a precipitate. Based on standard solubility rules, option 2 does produce a precipitate, while the others do not. However, the analysis here focuses on evaluating each option rather than selecting one; consider how solubility rules lead to the presence or absence of a solid phase in each case.

Which combination will result is a precipitate forming?

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