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JEE MAIN - Chemistry (2021 - 16th March Evening Shift - No. 21)

When 35 mL of 0.15 M lead nitrate solution is mixed with 20 mL of 0.12 M chromic sulphate solution, _________ $$\times$$ 10$$-$$5 moles of lead sulphate precipitate out. (Round off to the Nearest Integer).
Answer
525

Explanation

JEE Main 2021 (Online) 16th March Evening Shift Chemistry - Some Basic Concepts of Chemistry Question 135 English Explanation 1

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)

JEE Main 2021 (Online) 16th March Evening Shift Chemistry - Some Basic Concepts of Chemistry Question 135 English Explanation 2

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

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

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