QJEEM21Ch6edc: JEE Main 2021 Chemistry - Class 11 JEEM21Ch6edc Solution: When 35 mL of 0.15 M lead nitrate solution is mixed with 20 mL…

For 3 moles of Pb(NO₃)₂ , we require 1 mole of Cr₂(SO₄)₃

For 5.25 moles of Pb(NO₃)₂, we require $$\frac{1}{3}\times 5.25$$ mole of Cr₂(SO₄)₃ = 1.75 moles

But we have 2.4 moles. So, Cr₂(SO₄)₃ is excess reagent and Pb(NO₃)₂ is limiting reagent, (LR)

Moles of PbSO₄ formed = moles of Pb(NO₃)₂ consumed

= 5.25 m mol = 525 $$\times$$ 10^-5 moles