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Grade 11Organic Chemistry

The volume of O2 liberate from 0.96 g of H2O2 at STP

Yuvrajsinghyadav

8 Years agoGrade 11

2 Answers

Arun

8 Years ago

Dear student

no.of moles of H2O2 is= 0.96/68=0.014

(68=molecular mass of hydrogen peroxide)
H2O2===>H2 + O2
hence volume of O2 evolved =22.4L*0.014=0.3136L=313.6 mL (22.4L=volume of 1 mole of ideal gas)

Regards

Arun (askIITians forum expert)

Kushagra Madhukar

6 Years ago

Dear student,

Please find the attached solution to your question.

No.of moles of H₂O₂ = 0.96/68 = 0.014

(molecular mass of hydrogen peroxide = 68)
H₂O₂ → H₂ + O₂
Hence, Volume of O₂ evolved = 22.4L*0.014 = 0.3136L = 313.6 mL (volume of 1 mole of ideal gas = 22.4L)

Hope it helps.

Thanks and regards,

Kushagra

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