Question
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
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
Answers
GPT-4o mini
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.
\[
2 \text{H}_2\text{O}_2 (aq) \rightarrow 2 \text{H}_2\text{O} (l) + \text{O}_2 (g)
\]
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.