When 35 mL of 0-15 M lead nitrate solution is mixed with 20 mL of 0-12 M chromic sulphate solution- - 10-5 moles of lead sulphate precipitate out- Round off to the nearest integer-

3Pb(NO_3)_2 + Cr_2(SO_4)_3 → 3PbSO_4 ↓ + 2Cr(NO_3)_3 35× 0.15=5.25 mmol 20× 0.12=2.4mmol Moles of PbSO_4 = moles o ...

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Standard XII

Chemistry

Balancing Redox Reaction

When 35 mL of...

Question

When $35 m L$ 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).

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Solution

$3 P b (N O_{3})_{2} + C r_{2} (S O_{4})_{3} \to 3 P b S O_{4} ↓ + 2 C r (N O_{3})_{3} 35 \times 0.15 = 5.25 mmol 20 \times 0.12 = 2.4 mmol$

Moles of $P b S O_{4}$ = moles of $P b (N O_{3})_{2}$
$= 5.25 mmol = 525 \times 10^{- 5} mol$

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When 35 mL of 0.15 M lead nitrate solution is mixed with 20 mL of 0.12 M chromic sulphate solution, ×10−5 moles of lead sulphate precipitate out. (Round off to the nearest integer).

3Pb(NO3)2 + Cr2(SO4)3 →3PbSO4↓+2Cr(NO3)335×0.15=5.25 mmol 20×0.12=2.4mmol Moles of PbSO4 = moles of Pb(NO3)2 =5.25 mmol=525 ×10−5mol

When $35 m L$ 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).

$3 P b (N O_{3})_{2} + C r_{2} (S O_{4})_{3} \to 3 P b S O_{4} ↓ + 2 C r (N O_{3})_{3} 35 \times 0.15 = 5.25 mmol 20 \times 0.12 = 2.4 mmol$

Moles of $P b S O_{4}$ = moles of $P b (N O_{3})_{2}$
$= 5.25 mmol = 525 \times 10^{- 5} mol$