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Difficulty: MediumAmmonia and Trioxonitrate(V) Acid Preparation and Reactions
During the catalytic oxidation step in the Ostwald process for the industrial preparation of trioxonitrate(V) acid, ammonia gas reacts with excess oxygen according to the reaction equation:
4NH3(g)+5O2(g)→4NO(g)+6H2O(g)4\text{NH}_3(g) + 5\text{O}_2(g) \rightarrow 4\text{NO}(g) + 6\text{H}_2\text{O}(g)
What is the volume of nitrogen(II) oxide gas (NO\text{NO}) produced at STP when 17.0 g17.0\text{ g} of ammonia gas is completely oxidized?
(Molar mass of NH3=17.0 g mol−1\text{NH}_3 = 17.0\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
    24.0 dm324.0\text{ dm}^3
  2. 22.4 dm322.4\text{ dm}^3Answer
  3. C
    5.6 dm35.6\text{ dm}^3
  4. D
    89.6 dm389.6\text{ dm}^3

Answer

22.4 dm322.4\text{ dm}^3
One mole of ammonia (17.0 g17.0\text{ g}) yields exactly one mole of nitrogen(II) oxide gas according to the 1:1 mole ratio in the balanced chemical equation. At standard temperature and pressure (STP), one mole of any ideal gas occupies 22.4 dm322.4\text{ dm}^3. Therefore, the volume of nitrogen(II) oxide gas produced is 22.4 dm322.4\text{ dm}^3.

Step-by-Step Solution

1
Calculate the amount of ammonia gas in moles.
Moles of NH3=17.0 g17.0 g mol−1=1.0 mol\text{Moles of NH}_3 = \frac{17.0\text{ g}}{17.0\text{ g mol}^{-1}} = 1.0\text{ mol}
Dividing given mass by molar mass determines the number of moles of reactant present.
2
Determine the mole ratio between ammonia (NH3\text{NH}_3) and nitrogen(II) oxide (NO\text{NO}).
Mole ratio of NH3:NO=4:4=1:1\text{Mole ratio of NH}_3 : \text{NO} = 4 : 4 = 1 : 1. Therefore, 1.0 mol1.0\text{ mol} of NH3\text{NH}_3 produces 1.0 mol1.0\text{ mol} of NO\text{NO}.
The balanced chemical equation shows that 4 moles of NH3\text{NH}_3 produce 4 moles of NO\text{NO}.
3
Calculate the volume of nitrogen(II) oxide gas produced at STP.
Volume of NO=1.0 mol×22.4 dm3 mol−1=22.4 dm3\text{Volume of NO} = 1.0\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 22.4\text{ dm}^3
Multiplying moles of product by the molar gas volume at STP (22.4 dm3 mol−122.4\text{ dm}^3\text{ mol}^{-1}) yields the total gas volume.

Key Concept

Gas Stoichiometry and Ostwald Process Oxidation
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Topics

Ammonia and Trioxonitrate(V) Acid Preparation and Reactions

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