For the reaction represented by the equation pb(no3)2 + 2ki → pbi2 + 2kno3, how many moles of lead(ii) iodide are produced from 300. g of potassium iodide and an excess of pb(no3)2? a. 11.0 mol selected: b. 1.81 mol c. 3.61 mol d. 0.904 mol

Number of moles equals of potassium iodide;

The molar mass of potassium iodide is 166 g/mole

Moles = 300/166

            = 1.8072moles

According to the equation;

2 moles of KI produces 1 mole of PbI2 9lead (ii) iodide

Therefore; the number of moles of lead (ii) iodide produced will be;

                = 1.8072/2

                = 0.9036moles

Thus the number of moles of lead (ii) iodide is 0.904 mole

tax

Explanation: