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Most Important Selected Qs for JEE AdvancedChemistryRedox Reactions

H ₂ O ₂ is reduced rapidly by Sn ²⁺ , the products being Sn ⁴⁺ and water. H ₂ O ₂ decomposes slowly at room temperature to yield O ₂ and water. Calculate the volume of O ₂ produced at 20^ C and 1 atm . when 200 gm . of 10 % by mass H ₂ O ₂ in water is treated with 100 ml . of 2 M Sn ²⁺ and then the mixture is allowed to stand until no further reaction occurs. (mark the answer in L)

Correct answer

4.66

Step-by-step solution

Equivalents of H ₂ O ₂ intially = 200 10 100 1 34 2=1.176 Equivalents of Sn ²⁺ = 2 2 100 1000 =0.4 Equivalents of H ₂ O ₂ left =1.176-0.4=0.776 Moles of H ₂ O ₂ left = 0.776 2 =0.388 Moles of O ₂ produced = 0.388 2 [ H ₂ O ₂ H ₂ O + 1 2 O ₂ ]=0.194 Volume of O ₂ = 0.194 0.082 293 1 =4.66 ~L

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