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.