A solution containing excess lead(ii) nitrate is reacted with 380.0ML of 0.250mol/L potassium iodide solution. A bright yellow Precipitate is of lead(ii) iodide is formed.

Question

Calculate the mass of lead (ii) iodide that should be formed.

Pb(NO3)2+2KI-> 2KNO3+PbI2

Answer 1

mols KI = M x L = ?
Use the coefficients in the balanced equation to convert mols KI to mols PbI2.
Now convert mols PbI2 to grams. g = mols x molar mass = ?