the decomposition of hydrogen peroxide in the presence of a catalyst at standard pressure and room temperature is given by the unbalanced chemical equation below.

the oxygen gas is collected and the volume is recorded over a period of time the reaction is completed at time t.
H_2O_2 (aq) - >H_2O (litre) +O_2(g)
Take the molar gas volume (Vm) as 24,45 dm^3 at room temperature and standard temperature.
use the information above to calculate the mass of hydrogen peroxide that decomposed

1 answer

To calculate the mass of hydrogen peroxide (H₂O₂) that decomposed, we need to first balance the chemical equation for the decomposition of hydrogen peroxide:

\[ 2 \text{H}_2\text{O}_2 (aq) \rightarrow 2 \text{H}_2\text{O} (l) + \text{O}_2 (g) \]

From the balanced equation, we can see that 2 moles of H₂O₂ produce 1 mole of O₂ gas.

Next, we need to determine the volume of oxygen gas (O₂) collected during the reaction at room temperature and standard pressure. Let's denote this volume as \( V_{\text{O}_2} \) in dm³.

Using the molar gas volume (Vm) at room temperature and standard pressure, which is given as 24.45 dm³, we can find the number of moles of O₂ produced:

\[ \text{Moles of } O_2 = \frac{V_{\text{O}_2}}{V_m} \]

This means:

\[ n_{O_2} = \frac{V_{\text{O}_2}}{24.45 , \text{dm}^3} \]

According to the stoichiometry of the balanced reaction, the relationship between moles of H₂O₂ and moles of O₂ is as follows:

\[ \text{Moles of } H_2O_2 = 2 \times \text{Moles of } O_2 \]

Thus, we can write:

\[ n_{H_2O_2} = 2 \left(\frac{V_{\text{O}_2}}{24.45}\right) \]

Next, we need to calculate the mass of hydrogen peroxide that decomposed. To do this, we will use the molar mass of hydrogen peroxide (H₂O₂). The molar mass of H₂O₂ is calculated as follows:

  • H (Hydrogen): 1 g/mol × 2 = 2 g/mol
  • O (Oxygen): 16 g/mol × 2 = 32 g/mol

Therefore, the molar mass of H₂O₂ is: \[ \text{Molar mass of } H_2O_2 = 2 + 32 = 34 , \text{g/mol} \]

Now, we can calculate the mass of H₂O₂:

\[ \text{Mass of } H_2O_2 = n_{H_2O_2} \times \text{molar mass of } H_2O_2 \]

Substituting \( n_{H_2O_2} \):

\[ \text{Mass of } H_2O_2 = 2 \left(\frac{V_{\text{O}_2}}{24.45}\right) \times 34 \]

This gives:

\[ \text{Mass of } H_2O_2 = \frac{68 V_{\text{O}_2}}{24.45} \]

So, if you have a specific volume of oxygen gas collected \( V_{\text{O}_2} \) in dm³, you can substitute that into the equation to find the mass of hydrogen peroxide that decomposed.

For example, if \( V_{\text{O}_2} = 1 , \text{dm}^3 \):

\[ \text{Mass of } H_2O_2 = \frac{68 \times 1}{24.45} \approx 2.78 , \text{g} \]

You would follow this calculation for the actual volume measured during the experiment.