Molar mass of H2O2=34 gH_2O_2= 34 \ gH2O2=34 g
Moles of H2O2=434H_2O_2= \frac{4}{34}H2O2=344
2H2O2→2H2O+O22H_2O_2\to 2H_2O + O_22H2O2→2H2O+O2
So ,moles of oxygen produced will be half of hydrogen peroxide .
n=234n= \frac{2}{34}n=342
T=500K,V=250 ml=0.25 LT=500K,V= 250\ ml = 0.25 \ LT=500K,V=250 ml=0.25 L
Using ideal gas equation,
PV=nRTPV= nRTPV=nRT
P×0.25=234×0.0821×500 ⟹ P\times 0.25= \frac{2}{34}\times 0.0821 \times 500\impliesP×0.25=342×0.0821×500⟹
P=9.66 atmP=9.66 \ atmP=9.66 atm
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