What volume of pure , collected at 25°C and 737 torr, would be generated by decomposition of 123 g of a 50.0% by mass hydrogen peroxide solution? Ignore any water vapor that may be present.
The correct answer and explanation is:
To find the volume of oxygen gas (O₂) generated by the decomposition of hydrogen peroxide (H₂O₂), we need to follow these steps:
Step 1: Write the balanced chemical equation
2H2O2(aq)→2H2O(l)+O2(g)2 \text{H}_2\text{O}_2 (aq) \rightarrow 2 \text{H}_2\text{O} (l) + \text{O}_2 (g)
This tells us that 2 moles of H₂O₂ produce 1 mole of O₂ gas.
Step 2: Determine moles of H₂O₂ in the sample
You are given:
- Mass of solution = 123 g
- H₂O₂ concentration = 50.0% by mass
Mass of H2O2=0.50×123 g=61.5 g\text{Mass of H}_2\text{O}_2 = 0.50 \times 123 \, \text{g} = 61.5 \, \text{g}
Molar mass of H₂O₂ = 34.02 g/mol Moles of H2O2=61.534.02≈1.808 mol\text{Moles of H}_2\text{O}_2 = \frac{61.5}{34.02} \approx 1.808 \, \text{mol}
Step 3: Use stoichiometry to find moles of O₂
From the equation: 2mol H2O2→1mol O22 \text{mol H}_2\text{O}_2 \rightarrow 1 \text{mol O}_2 Moles of O2=1.8082=0.904 mol\text{Moles of O}_2 = \frac{1.808}{2} = 0.904 \, \text{mol}
Step 4: Use the Ideal Gas Law to find volume
The Ideal Gas Law: PV=nRT⇒V=nRTPPV = nRT \quad \Rightarrow \quad V = \frac{nRT}{P}
Given:
- n=0.904 moln = 0.904 \, \text{mol}
- T=25∘C=298 KT = 25^\circ C = 298 \, \text{K}
- P=737 torr=737760 atm≈0.970 atmP = 737 \, \text{torr} = \frac{737}{760} \, \text{atm} \approx 0.970 \, \text{atm}
- R=0.0821 L⋅atmmol⋅KR = 0.0821 \, \frac{L \cdot atm}{mol \cdot K}
V=(0.904)(0.0821)(298)0.970≈22.0660.970≈22.74 LV = \frac{(0.904)(0.0821)(298)}{0.970} \approx \frac{22.066}{0.970} \approx 22.74 \, \text{L}
✅ Final Answer:
22.7 L of O2\boxed{22.7 \, \text{L of O}_2}
Explanation (Summary)
Hydrogen peroxide decomposes into water and oxygen gas. Given a 50% H₂O₂ solution weighing 123 g, we first find the actual mass and moles of H₂O₂. Using stoichiometry from the balanced equation, we determine the moles of O₂ gas produced. Finally, we apply the Ideal Gas Law to calculate the volume of oxygen gas at the specified conditions. The result is 22.7 liters of oxygen gas at 25°C and 737 torr.