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Difficulty: EasyTetraoxosulfate(VI) Acid: Contact Process and Properties

In the Contact Process for the industrial manufacture of tetraoxosulfate(VI) acid, sulfur(IV) oxide gas reacts with oxygen gas according to the equation: 2SO2(g)+O2(g)→2SO3(g)2SO_2(g) + O_2(g) \rightarrow 2SO_3(g). What volume of oxygen gas, measured in dm3\text{dm}^3 at stp, is required to completely react with 56 dm356\text{ dm}^3 of SO2SO_2 gas at stp?

Answer: 28 dm³

Answer

The volume of oxygen gas required at stp is 28 dm³.
According to the balanced chemical equation 2SO2(g)+O2(g)→2SO3(g)2SO_2(g) + O_2(g) \rightarrow 2SO_3(g), 2 volumes of SO2SO_2 gas require 1 volume of O2O_2 gas for complete oxidation. Therefore, 56 dm356\text{ dm}^3 of SO2SO_2 requires half its volume in oxygen, which equals 28 dm328\text{ dm}^3.

Step-by-Step Solution

1
Determine the stoichiometric ratio between SO2SO_2 and O2O_2 from the balanced equation.
2 volumes of SO2(g)SO_2(g) react with 1 volume of O2(g)O_2(g).
By Gay-Lussac's Law of Combining Volumes, gases react in simple whole-number volume ratios under the same conditions of temperature and pressure.
2
Compute the required volume of O2O_2 gas for 56 dm356\text{ dm}^3 of SO2SO_2.
Volume of O2=56 dm32=28 dm3\text{Volume of } O_2 = \frac{56\text{ dm}^3}{2} = 28\text{ dm}^3.
Since the ratio of SO2SO_2 to O2O_2 is 2:12:1, the volume of oxygen gas required is half the volume of sulfur(IV) oxide gas.

Key Concept

Stoichiometric Volume Calculations in Gas Reactions (Contact Process)
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Topics

Tetraoxosulfate(VI) Acid: Contact Process and Properties

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