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 decomposes
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 decomposes
Answers
GPT-4o mini
To calculate the mass of hydrogen peroxide (H₂O₂) that decomposes, we first need to balance the given chemical equation for the decomposition of hydrogen peroxide. The balanced equation is:
\[ 2 \, \text{H}_2\text{O}_2(aq) \rightarrow 2 \, \text{H}_2\text{O}(l) + \text{O}_2(g) \]
This equation tells us that 2 moles of hydrogen peroxide decompose to produce 2 moles of water and 1 mole of oxygen gas.
Let's say the volume of oxygen gas (O₂) collected at standard temperature and pressure (STP) over the time \( t \) is \( V \) dm³.
Using the molar gas volume \( V_m = 24.45 \, \text{dm}^3/\text{mol} \), we can find the number of moles of oxygen gas produced:
\[
\text{Moles of } O_2 = \frac{V}{V_m} = \frac{V}{24.45 \, \text{dm}^3/\text{mol}}
\]
From the balanced equation, we see that for every 1 mole of oxygen gas produced, 2 moles of hydrogen peroxide are decomposed. Therefore, the moles of hydrogen peroxide that decomposes can be calculated as follows:
\[
\text{Moles of } H_2O_2 = 2 \times \text{Moles of } O_2 = 2 \left(\frac{V}{24.45}\right) = \frac{2V}{24.45}
\]
Next, we need to find the mass of hydrogen peroxide that decomposes. The molar mass of hydrogen peroxide (H₂O₂) is calculated as follows:
- Hydrogen (H): 1 g/mol × 2 = 2 g/mol
- Oxygen (O): 16 g/mol × 2 = 32 g/mol
The total 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 hydrogen peroxide that decomposes:
\[
\text{Mass of } H_2O_2 = \text{Moles of } H_2O_2 \times \text{Molar mass of } H_2O_2
\]
Substituting in our expression for moles:
\[
\text{Mass of } H_2O_2 = \left(\frac{2V}{24.45}\right) \times 34
\]
Thus, simplifying gives:
\[
\text{Mass of } H_2O_2 = \frac{68V}{24.45} \, \text{g}
\]
You will need the specific volume \( V \) in dm³ to get a numerical answer. Simply substitute the collected volume of oxygen gas into the equation to find the mass of hydrogen peroxide that decomposed.
\[ 2 \, \text{H}_2\text{O}_2(aq) \rightarrow 2 \, \text{H}_2\text{O}(l) + \text{O}_2(g) \]
This equation tells us that 2 moles of hydrogen peroxide decompose to produce 2 moles of water and 1 mole of oxygen gas.
Let's say the volume of oxygen gas (O₂) collected at standard temperature and pressure (STP) over the time \( t \) is \( V \) dm³.
Using the molar gas volume \( V_m = 24.45 \, \text{dm}^3/\text{mol} \), we can find the number of moles of oxygen gas produced:
\[
\text{Moles of } O_2 = \frac{V}{V_m} = \frac{V}{24.45 \, \text{dm}^3/\text{mol}}
\]
From the balanced equation, we see that for every 1 mole of oxygen gas produced, 2 moles of hydrogen peroxide are decomposed. Therefore, the moles of hydrogen peroxide that decomposes can be calculated as follows:
\[
\text{Moles of } H_2O_2 = 2 \times \text{Moles of } O_2 = 2 \left(\frac{V}{24.45}\right) = \frac{2V}{24.45}
\]
Next, we need to find the mass of hydrogen peroxide that decomposes. The molar mass of hydrogen peroxide (H₂O₂) is calculated as follows:
- Hydrogen (H): 1 g/mol × 2 = 2 g/mol
- Oxygen (O): 16 g/mol × 2 = 32 g/mol
The total 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 hydrogen peroxide that decomposes:
\[
\text{Mass of } H_2O_2 = \text{Moles of } H_2O_2 \times \text{Molar mass of } H_2O_2
\]
Substituting in our expression for moles:
\[
\text{Mass of } H_2O_2 = \left(\frac{2V}{24.45}\right) \times 34
\]
Thus, simplifying gives:
\[
\text{Mass of } H_2O_2 = \frac{68V}{24.45} \, \text{g}
\]
You will need the specific volume \( V \) in dm³ to get a numerical answer. Simply substitute the collected volume of oxygen gas into the equation to find the mass of hydrogen peroxide that decomposed.