Question

For each of the following precipitation reactions, calculate how many grams of the first reactant are necessary to completely react with 54.4 g of the second reactant. 2KI(aq)+Pb(NO3)2(aq)→PbI2(s)+2KNO3(aq)

Answer 1

Mass of Pb(NO₃)₂ = 54.4 g

Molar mass of Pb(NO₃)₂ = 331.2 g/mol

Moles of Pb(NO₃)₂ = Mass of Pb(NO₃)₂ / Molar mass of Pb(NO₃)₂

                                  = 54.4/331.2 = 0.1642 moles

2KI(aq) + Pb(NO₃)₂(aq) PbI₂(s) + 2KNO₃(aq)

From the balanced reaction, one mole of Pb(NO₃)₂ react with two moles of KI

Therefore, moles of KI required for complete reaction = 2 x 0.1642 = 0.3285 moles

Mass of KI required = Moles of KI x Molar mass of KI = 0.3285 x 166 g/mol = 54.53 g

Mass of KI necessary to completely react with 54.4 g of Pb(NO₃)₂ = 54.53 g