Question

How Can We Determine the Actual Percentage of H2O2 in a Drugstore Bottle of Hydrogen Peroxide?

Answer 1

Measuring the volume of oxygen released

Step-by-step explanation:

The hydrogen peroxide has a decomposition reaction:

`2 H_2O_2 \longrightarrow H_2O + O_2`

So, if we leave this reaction happen and we capture the oxygen released measuring its volume we can determine the percentage of peroxide.

`n_(O2)=(P*V_(O2))/(T*R)`

`n_(H2O2)= 0.5*n_(O2)=0.5* (P*V_(O2))/(T*R)`

`m_(H2O2)=n_(H2O2)*M_(H2O2)=M_(H2O2)*0.5* (P*V_(O2))/(T*R)`

`V_(H2O2)=m_(H2O2)* \rho _(H2O2)`

`V_(H2O2)= \rho _(H2O2)*M_(H2O2)*0.5* (P*V_(O2))/(T*R)`

Where:

• 
  `n_(H2O2)` is the number of moles of H2O2
• 
  `m_(H2O2)` is mass of H2O2
• 
  `M_(H2O2)` is the molecular weight of H2O2
• 
  `\rho _(H2O2)` is the density of H2O2

`Percentage=(V_(H2O2))/(V_(total))*100`