Chem Stoichiometry multiple choice question

Question

How many moles of potassium iodide, KI, are required to precipitate all of the lead (II) ion from 25.0 mL of a 1.6 M Pb(NO₃)₂ solution?

a.) 0.64 mol

b.) 0.080 mole

c.) 0.040 mole

d.) 064 mol

e.) 0.020 mole

Answer 1

Great Day Areli!

Write balanced chemical equation

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

Calculate moles of Pb in the reaction: use molarity equation.

1.6 M = Mols of solute / .025 liters =.04 mols of Pb in the reaction.

Determine mole ratio of KI to Pb(NO3)2 by using the coefficients in the balanced chemical equation.

2 moles of KI = 1 mol Pb(NO3)2

.04 mol Pb ( 2 mol KI / 1 mol Pb(NO3)2) = .08 mol of KI required to precipitate all the Pb from solution.