Answer:
Li₂S(aq) + Pb(NO₃)₂(aq)
2 K₂S(aq) + Sn(NO₃)₄(aq)
Explanation:
For these reactions, the products are:
1. Na₂S(aq) + 2 KCl(aq) → K₂S + 2 NaCl
2. Li₂S(aq) + Pb(NO₃)₂(aq) → PbS + 2 Li(NO₃)
3. Pb(ClO₃)₂(aq) + 2 NaNO₃(aq) → Pb(NO₃)₂ + 2 Na(ClO₃)
4. AgNO₃(aq) + KCl(aq) → AgCl + KNO₃
5. 2 K₂S(aq) + Sn(NO₃)₄(aq) → SnS₂ + 4 KNO₃
K₂S is a soluble sulfide that, in solid state, is white. As K has a low electronegativity, the relative polar bond K-S allows its dissolution in water.
PbS is a black and insoluble compound. As Pb electronegativity is higher than K electronegativity, The bond is less-polar and its dissolution will not be allowed.
For the third reaction, all nitrates are soluble and Na(ClO₃) is also a very soluble compound.
AgCl is an insoluble white compound.
As Sn electronegativity is high, the solubility of this compound is very low. Also, this compound is black!
Thus, combinations could yield a black precipitate are:
Li₂S(aq) + Pb(NO₃)₂(aq)
2 K₂S(aq) + Sn(NO₃)₄(aq)
I hope it helps!
