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

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

Grade:11

2 Answers

Arun
25750 Points
6 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
askIITians Faculty 628 Points
3 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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