Answer to Question #308837 in Chemistry for tj woe

Question

what mass of lead (ii) iodide will be formed when 20.0 g of lead (ii) nitrate is added to 33.0g of potassium iodide? which is the limiting reactant?

Answer 1

lead(II) iodide = PbI₂

lead(II) nitrate = Pb(NO₃)₂

potassium iodide = KI

Solution:

The molar mass of KI is 166.0 g/mol

The molar mass of Pb(NO₃)₂ is 331.2 g/mol

Calculate the moles of each reactant:

Moles of Pb(NO₃)₂ = (20.0 g Pb(NO₃)₂) × (1 mol Pb(NO₃)₂ / 331.2 g Pb(NO₃)₂) = 0.0604 mol Pb(NO₃)₂

Moles of KI = (33.0 g KI) × (1 mol KI / 166.0 g KI) = 0.1988 mol KI

Balanced chemical equation:

Pb(NO₃)₂ + 2KI → PbI₂ + 2KNO₃

According to stoichiometry:

1 mol of Pb(NO₃)₂ reacts with 2 mol of KI

Thus, 0.0604 mol of Pb(NO₃)₂ reacts with:

(0.0604 mol Pb(NO₃)₂) × (2 mol KI / 1 mol Pb(NO₃)₂) = 0.1208 mol KI

However, initially there is 0.1988 mol of KI (according to the task).

Thus, Pb(NO₃)₂ acts as limiting reactant and KI is excess reactant.

According to stoichiometry:

1 mol of Pb(NO₃)₂ produces 1 mol of PbI₂

Thus, 0.0604 mol of Pb(NO₃)₂ produces:

(0.0604 mol Pb(NO₃)₂) × (1 mol PbI₂ / 1 mol Pb(NO₃)₂) = 0.0604 mol PbI₂

The molar mass of PbI₂ is 461.0 g/mol

Therefore,

Mass of PbI₂ = (0.0604 mol PbI₂) × (461.0 g PbI₂ / 1 mol PbI₂) = 27.8444 g PbI₂ = 27.8 g PbI₂

Mass of PbI₂ = 27.8 g

Answer:

The limiting reactant is Pb(NO₃)₂

27.8 grams of lead(II) iodide (PbI₂) will be formed