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.
