Question

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

Answer 1

Answer:

Measuring the volume of oxygen released

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.

$n_{O2}=\frac{P*V_{O2}}{T*R}$

$n_{H2O2}= 0.5*n_{O2}=0.5* \frac{P*V_{O2}}{T*R}$

$m_{H2O2}=n_{H2O2}*M_{H2O2}=M_{H2O2}*0.5* \frac{P*V_{O2}}{T*R}$

$V_{H2O2}=m_{H2O2}* \rho _{H2O2}$

$V_{H2O2}= \rho _{H2O2}*M_{H2O2}*0.5* \frac{P*V_{O2}}{T*R}$

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

$Percentage=\frac{V_{H2O2}}{V_{total}}*100$