Question

when you have mixed the solutions of lead(II) nitrate and potassium iodide.
a) what is the color of precipitate formed and also name the precipitate.
b)write the balanced chemical equation for this reaction.
c) Is this also a double displacement reaction.

Answer 1

Answer:

a. The precipitate is yellow in colour and the compound is lead (II) Iodide.

b. Pb (NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq).

c. Yes, it is a double displacement reaction.

Answer 2

Lead (II) Nitrate + Potassium Iodide → X + Y

Given condition is that Lead (II) Nitrate reacts with Potassium Iodide and give two compounds. Let them as X and Y.

When Lead (II) Nitrate reacts with Potassium Iodide formation of precipitate takes place.

The following reaction takes place:

$\sf{Pb(NO_3)_2\: (aq) + KI\: (aq) \rightarrow PbI_2 \:(s) + KNO_3 \:(aq)}$

a) When Lead (II) Nitrate reacts with Potassium Iodide formation Lead (II) Iodide which is a yellow coloured precipitate and Potassium Nitrate (formation) takes place.

b) Balanced chemical reaction is as follow:

$\sf{Pb(NO_3)_2\: (aq) + 2KI\: (aq) \rightarrow PbI_2 \:(s) + 2KNO_3 \:(aq)}$

c) Yes, it a Double Displacement Reaction. As the ions are displacing each other. Also it is a Precipitation Reaction as yellow precipitates of Lead (II) Iodine are formed.