Calculate the mass of precipitate that forms when 250.0 ml of an aqueous solution containing 35.5 g of lead(ii) nitrate reacts with excess sodium iodide solution.

= 49.42 g

The equation for the reaction between Lead (ii) nitrate and sodium iodide;

Pb(NO3)2(aq) + 2 NaI(aq) ---> 2NaNO3(aq) + PbI2(s)

The precipitate formed in this equation is Lead iodide

We first calculate the moles of lead nitrate;

Moles = mass/molar mass

           = 35.5 g/ 331.2 g/mol

           = 0.1072 moles

The mole ratio of Pb(NO3)2 : PbI2 is 1 : 1

Therefore; the number of moles of lead iodide is 0.1072 moles

Mass = moles × molar mass

         = 0.1072 moles × 461.01 g/mol

         = 49.42 g