Question

An aqueous solution containing 6.06 g of lead(II) nitrate is added to an aqueous solution containing 6.58 g of potassium chloride. Enter the balanced chemical equation for this reaction. Be sure to include all physical states. balanced chemical equation: What is the limiting reactant

Answer 1

Answer:

Pb(NO₃)₂ is the limiting reactant.

The equation is: Pb(NO₃)₂ (aq) + 2KCl (aq) →  PbCl₂ (s) ↓  + 2KNO₃ (aq)

Explanation:

We identify our reactants:

Pb(NO₃)₂ → Lead (II) nitrate

KCl → Potassium chloride

Our reaction is:

Pb(NO₃)₂ (aq)  +  2KCl (aq) →  PbCl₂ (s) ↓  + 2KNO₃ (aq)

1 mol of Lead (II) nitrate reacts to 2 moles of KCl, in order to produce 2 moles of potassium nitrate and 1 mol of slid Lead (II) chloride.

We determine the moles of the reactants:

6.06 g . 1mol /331.2 g = 0.0183 moles of Lead (II) nitrate

6.58 g . 1mol / 74.55g = 0.0882 moles of KCl

2 moles of KCl react to 1 mol of Pb(NO₃)₂

Then, 0.0882 moles of KCl may react to (0.0882 . 1) /2 = 0.0441 moles

We have 0.0183 moles of Pb(NO₃)₂ and we need 0.441 moles. Then, the

Pb(NO₃)₂ is our limiting reactant.