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Difficulty: MediumOxygen, Ozone, and Classification of Oxides
During the laboratory preparation of oxygen gas, a sample of 17.0 g17.0\text{ g} of hydrogen peroxide (H2O2\text{H}_2\text{O}_2) decomposes completely in the presence of a manganese(IV) oxide catalyst according to the equation:
2H2O2(aq)→MnO22H2O(l)+O2(g)2\text{H}_2\text{O}_2(\text{aq}) \xrightarrow{\text{MnO}_2} 2\text{H}_2\text{O}(\text{l}) + \text{O}_2(\text{g})
What volume of oxygen gas, measured at standard temperature and pressure (STP), is released in this process?
[Molar mass of H2O2=34.0 g mol−1\text{H}_2\text{O}_2 = 34.0\text{ g mol}^{-1}; Molar volume of gas at STP =22.4 dm3 mol−1= 22.4\text{ dm}^3\text{ mol}^{-1}]
  1. A
    2.80 dm32.80\text{ dm}^3
  2. 5.60 dm35.60\text{ dm}^3Answer
  3. C
    6.00 dm36.00\text{ dm}^3
  4. D
    11.2 dm311.2\text{ dm}^3

Answer

5.60 dm35.60\text{ dm}^3
The decomposition of 17.0 g17.0\text{ g} of H2O2\text{H}_2\text{O}_2 yields 0.50 mol0.50\text{ mol} of reactant. According to the balanced chemical equation, 2 moles2\text{ moles} of H2O2\text{H}_2\text{O}_2 yield 1 mole1\text{ mole} of O2\text{O}_2, producing 0.25 mol0.25\text{ mol} of oxygen gas. At standard temperature and pressure (STP), 0.25 mol0.25\text{ mol} occupies 0.25×22.4 dm3 mol−1=5.60 dm30.25 \times 22.4\text{ dm}^3\text{ mol}^{-1} = 5.60\text{ dm}^3.

Step-by-Step Solution

1
Calculate the amount in moles of hydrogen peroxide (H2O2\text{H}_2\text{O}_2) reactant.
n(H2O2)=17.0 g34.0 g mol−1=0.50 moln(\text{H}_2\text{O}_2) = \frac{17.0\text{ g}}{34.0\text{ g mol}^{-1}} = 0.50\text{ mol}
Converting mass to moles using the molar mass provides the quantity of reactant available.
2
Determine the moles of oxygen gas (O2\text{O}_2) formed using equation stoichiometry.
Since 2 mol H2O2→1 mol O22\text{ mol } \text{H}_2\text{O}_2 \rightarrow 1\text{ mol } \text{O}_2, n(O2)=0.50 mol2=0.25 moln(\text{O}_2) = \frac{0.50\text{ mol}}{2} = 0.25\text{ mol}
The balanced chemical equation shows a 2:1 molar ratio between reactant and gaseous product.
3
Calculate the volume of oxygen gas produced at STP.
V(O2)=0.25 mol×22.4 dm3 mol−1=5.60 dm3V(\text{O}_2) = 0.25\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 5.60\text{ dm}^3
Multiplying the calculated moles of gas by the standard molar gas volume gives the volume at STP.

Key Concept

Stoichiometric calculations and gas molar volume at STP for oxygen preparation.
Estimated Time:1m 30s
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Topics

Oxygen, Ozone, and Classification of Oxides

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