<h3>
Answer:</h3>
Li₂S(aq)+Pb(NO₃)₂(aq)
→ PbS(s) + LiNO₃(aq)
K2S(aq)+Sn(NO₃)₄(aq→ SnS(s) + KNO₃(aq)
<h3>
Explanation:</h3>
- According to solubility rules, metal sulfides are insoluble except Calcium sulfide(CaS), Magnesium sulfide(MgS), Barium sulfide(BaS), and those of potassium, sodium, and ammonium.
- Therefore, when all the other sulfides are formed during a reaction they form a precipitate that is shown to be in solid-state.
- Precipitates are formed during precipitation reactions when cations and anions combine to form a compound that is insoluble in water.
In our case; In the equations given, some reactions will take place to form a precipitate while others will not occur.
That is;
- Na₂S(aq)+KCl(aq) → No reaction
- Li₂S(aq)+Pb(NO₃)₂(aq) → PbS(s) + LiNO₃(aq)
- Pb(ClO₃)₂(aq)+NaNO₃(aq) → No reaction
- AgNO₃(aq)+KCl(aq) →AgCl(s) + KNO₃(aq)
- K2S(aq)+Sn(NO₃)₄(aq→ SnS(s) + KNO₃(aq)
The first reaction will not occur.
The second equation is a precipitation reaction that forms lead(ii)sulfide which is a black precipitate.
The third reaction will not take place.
The fourth reaction will be a precipitation reaction forming silver chloride precipitate.
The fifth equation is also a precipitation reaction that forms tin(ii) sulfide, SnS which is a black precipitate.
Therefore; the equations that answers our question are;
- Li₂S(aq)+Pb(NO₃)₂(aq) → PbS(s) + LiNO₃(aq)
- K2S(aq)+Sn(NO₃)₄(aq→ SnS(s) + KNO₃(aq)
