Answer to Question #313599 in General Chemistry for Ivan Ackerman

Question

An aqueous solution containing 7.11 g

7.11 g of lead(II) nitrate is added to an aqueous solution containing 6.53 g

6.53 g of potassium chloride.

Enter the balanced chemical equation for this reaction. Be sure to include all physical states. What is the limiting reactant?

Answer 1

Solution:

lead(II) nitrate = Pb(NO₃)₂

potassium chloride = KCl

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

The molar mass of KCl is 74.55 g/mol

Calculate moles of each reactant:

Moles of Pb(NO₃)₂ = (7.11 g Pb(NO₃)₂) × (1 mol Pb(NO₃)₂ / 331.2 g Pb(NO₃)₂) = 0.0215 mol Pb(NO₃)₂

Moles of KCl = (6.53 g g KCl) × (1 mol KCl / 74.55 g KCl) = 0.0876 mol KCl

Balanced chemical equation:

Pb(NO₃)₂(aq) + 2KCl(aq) ⟶ PbCl₂(s) + 2KNO₃(aq)

According to stoichiometry:

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

Thus, 0.0215 mol of Pb(NO₃)₂ reacts with:

(0.0215 mol Pb(NO₃)₂) × (2 mol KCl / 1 mol Pb(NO₃)₂) = 0.0430 mol KCl

However, initially there is 0.0876 mol of KCl (according to the task).

Therefore, Pb(NO₃)₂ acts as limiting reactant and KCl is excess reactant.

Answer:

1) Balanced chemical equation: Pb(NO₃)₂(aq) + 2KCl(aq) ⟶ PbCl₂(s) + 2KNO₃(aq)

2) The limiting reactant is lead(II) nitrate - Pb(NO₃)₂