Question

Rewrite the following as an unbalanced chemical equation (do include phases of matter, you may look up the solubility chart):

When a student adds 20 drops of a solution of lead(II) nitrate to a solution of magnesium chloride, a white precipitate (powder) of lead(II) chloride is observed. The student correctly concludes that the other product must be magnesium nitrate.
A) Pb(NO3)2 (aq) + MgCl2 (aq) -> PbCl2 (s) + Mg(NO3)2 (aq)
B) PbCl2 (s) + Mg(NO3)2 (aq) -> Pb(NO3)2 (aq) + MgCl2 (aq)
C) Pb(NO3)2 (s) + MgCl2 (aq) -> PbCl2 (aq) + Mg(NO3)2 (s)
D) PbCl2 (aq) + Mg(NO3)2 (s) -> Pb(NO3)2 (s) + MgCl2 (aq)

Answer 1

The unbalanced chemical equation for the reaction between lead(II) nitrate and magnesium chloride is Pb(NO3)2 (aq) + MgCl2 (aq) → PbCl2 (s) + Mg(NO3)2 (aq).

Step-by-step explanation:

The unbalanced chemical equation for the reaction between lead(II) nitrate and magnesium chloride is:

Pb(NO3)2 (aq) + MgCl2 (aq) → PbCl2 (s) + Mg(NO3)2 (aq)

When the two solutions react, a white precipitate of lead(II) chloride is formed, indicating the formation of a solid. The correct conclusion drawn by the student is that the other product is magnesium nitrate, which remains in solution.