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.

Answer;

= 49.42 g

Explanation;

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