Answer:
AgNO₃(aq) + NaCl(aq) ⇄ AgCl(s)↓ + NaNO₃(aq) Kps
2K₃PO₄(aq) + 3MgCl₂(aq) ⇄ 6KCl (aq) + Mg₃(PO₄)₂ (s)↓ Kps
Explanation:
In the first reaction (1) we have:
Silver nitrate and sodium chloride as the reactants. Then the products will be, silver chloride (s) and sodium nitrate.
We know that all the salts from the nitrate are soluble, so the chloride will react to the silver cation to make the precipitate, an insoluble salt.
AgNO₃(aq) + NaCl(aq) ⇄ AgCl(s)↓ + NaNO₃(aq) Kps
For the second reaction (2), the reactants are:
Potassium phosphate and magnessium chloride
Phosphate salts are insoluble, except for the group 1. The same as chloride that makes soluble salts with the elements from the group 1.
The equation that makes the precipitate is.
2K₃PO₄(aq) + 3MgCl₂(aq) ⇄ 6KCl (aq) + Mg₃(PO₄)₂ (s)↓ Kps
Answer:
1) AgNO3(aq) + NaCl(aq) → NaNO3(aq) + AgCl(s)↓
2) 2K3PO4(aq) + 3MgCl2(aq) → Mg3(PO4)2(s)↓ + 6KCl(aq)
Explanation:
1) AgNO3(aq)+NaCl(aq) <------>
Step 1: The unbalanced equation
AgNO3(aq)+NaCl(aq) → Ag+ + NO3- + Na+ + Cl-
Step 2: The balanced equation
AgNO3(aq) + NaCl(aq) → NaNO3(aq) + AgCl(s)↓
2) K3PO4(aq)+MgCl2(aq) <------->
Step 1: The unbalanced equation
K3PO4(aq) + MgCl2(aq) → 3K+ + PO4^3- + Mg^2+ + 2Cl-
K3PO4(aq) + MgCl2(aq) → Mg3(PO4)2(s) + KCl(aq)
Step 2: The balanced equation
2K3PO4(aq) + 3MgCl2(aq) → Mg3(PO4)2(s)↓ + 6KCl(aq)
