Answer:
1. Zn(OH)₂ (s)
2. Ag₂CO₃ (s)
3. Ni₃(PO₄)₂(s)
4. No reaction
5. (NH₄)₂CO₃(s)
Explanation:
Let's state the equations and we analyse some solubility and precipitation information:
ZnCl₂(aq) + 2KOH(aq) → Zn(OH)₂ (s) + 2KCl (aq)
All the salts from the halogens with group 1, are soluble.
The OH⁻ reacts to Zn cation in order to produce a precipitate. This is ok, but if the base is in excess, the Zn(OH)₂ will be soluble
K₂CO₃(aq) + 2AgNO₃(aq) → Ag₂CO₃ (s) ↓ + 2KNO₃(aq)
All salts from nitrate are soluble
All salts from carbonates are insoluble
2(NH₄)₃PO₄(aq) + 3Ni(NO₃)₂(aq) → Ni₃(PO₄)₂(s) ↓ + 6NH₄NO₃(aq)
Salts from phosphates are insoluble
All salts from nitrate are soluble
NaCl(aq) + KNO3(aq) → NO REACTION
All salts from nitrate are soluble
All the salts from the halogens with group 1, are soluble
Na₂CO₃(aq) + 2NH₄Cl(aq) → 2NaCl(aq) + (NH₄)₂CO₃(s) ↓
All salts from carbonates are insoluble
All the salts from the halogens with group 1, are soluble
Answer:
1) Zn(OH)2(s)
2)Ag2CO3(s)
3)Ni3(PO4)2(s)
4)/
5)(NH4)2CO3(s)
Explanation:
ZnCl2(aq) + 2KOH(aq)→ Zn^2+ +2Cl- + 2K+ + 2OH-
ZnCl2(aq) + 2KOH(aq)→ Zn(OH)2(s) + 2KCl(aq)
K2CO3(aq) + 2AgNO3(aq)→ 2K+ + CO3^2- + 2Ag+ + 2NO3-
K2CO3(aq) + 2AgNO3(aq)→ 2KNO3(aq) + Ag2CO3(s)
2(NH4)3PO4(aq) + 3Ni(NO3)2(aq) → 6NH4+ + 2PO4^3- + 3NI^2+ + 6NO3-
2(NH4)3PO4(aq) + 3Ni(NO3)2(aq) → Ni3(PO4)2(s) + 6NH4NO3(aq)
NaCl(aq) + KNO3(aq)→ Na+ + Cl- + K+ + NO3-
NaCl(aq) + KNO3(aq)→ NaNO3(aq) + KCl(aq)
Na2CO3(aq) + 2NH4Cl(aq)→ 2Na+ + CO3^2- + 2NH4+ + 2Cl-
Na2CO3(aq) + 2NH4Cl(aq)→ 2NaCl(aq) +( NH4)2CO3(s)
