For the following: AgNO3 (aq) + Na2S (aq) –> , assume a metathesis reaction (“5 step process”) and write the products.” NaAg Na2NO3 and AgS NaS and AgN NaNO3 and Ag2S “For the following: AgNO3 (aq) + Na2S (aq) –> , assume a metathesis reaction (“5 step process”) and identify the spectator ion(s).” aqueous Ag+ only aqueous Na+ and NO3- aqueous S-2 and Na+ aqueous S-2 only “For the following: AgNO3 (aq) + Na2S (aq) –> , assume a metathesis reaction (“5 step process”) and write the net ionic equation.” 2Ag+(aq) + S-2(aq) –> Ag2S(s) 2NO+(aq) + Cl-2(aq) –> ClNO(s) 2NO+(aq) + Cl-2(aq) –> ClNO(s) NO-1(aq) –> NO2(s)

Correct Answers:

1. Products of the reaction:
NaNO3 and Ag2S

2. Spectator ions:
Aqueous Na⁺ and NO₃⁻

3. Net ionic equation:
2Ag⁺(aq) + S²⁻(aq) → Ag₂S(s)


Explanation

This reaction is a classic example of a metathesis reaction, also known as a double replacement reaction. These reactions typically occur in aqueous solution and involve the exchange of ions between two compounds. The goal is to form a precipitate, gas, or weak electrolyte.

Let us go through the “5 step process” for metathesis reactions:

1. Write the formulas of the reactants:
AgNO₃ (silver nitrate) and Na₂S (sodium sulfide) are both soluble in water and thus exist as ions:
AgNO₃ → Ag⁺ + NO₃⁻
Na₂S → 2Na⁺ + S²⁻

2. Predict the products by switching the cations:
Ag⁺ pairs with S²⁻
Na⁺ pairs with NO₃⁻
This yields Ag₂S and NaNO₃

3. Determine solubility of the products:
Ag₂S is insoluble in water and will precipitate out of solution
NaNO₃ is soluble in water and remains in aqueous form

4. Write the balanced molecular equation:
2AgNO₃(aq) + Na₂S(aq) → Ag₂S(s) + 2NaNO₃(aq)

5. Write the complete ionic equation and identify spectator ions:
2Ag⁺(aq) + 2NO₃⁻(aq) + 2Na⁺(aq) + S²⁻(aq) → Ag₂S(s) + 2Na⁺(aq) + 2NO₃⁻(aq)
Here, Na⁺ and NO₃⁻ ions appear on both sides and do not participate in the actual chemical change. They are spectator ions.

Net ionic equation:
2Ag⁺(aq) + S²⁻(aq) → Ag₂S(s)

This net ionic equation shows the actual chemical change – the formation of solid silver sulfide from aqueous ions. The key driving force here is the formation of an insoluble solid, which removes those ions from the solution.

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