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Difficulty: HardTetraoxosulfate(VI) Acid: Contact Process and Properties
In an industrial Contact Process plant, sulfur(IV) oxide (SO2SO_2) gas is catalytically oxidized to sulfur(VI) oxide (SO3SO_3) according to the equation:
2SO2(g)+O2(g)⇌2SO3(g)2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)
If 67.2 dm367.2\text{ dm}^3 of SO2SO_2 measured at STP is reacted with excess oxygen gas, and the reaction achieves a 90%90\% conversion yield of SO3SO_3, what is the mass in grams of tetraoxosulfate(VI) acid (H2SO4H_2SO_4) produced when all the formed SO3SO_3 is absorbed in concentrated H2SO4H_2SO_4 and subsequently diluted with water? (Molar mass of H2SO4=98 g/molH_2SO_4 = 98\text{ g/mol}, molar volume of gas at STP =22.4 dm3/mol= 22.4\text{ dm}^3\text{/mol})

Answer: 264.6 g

Answer

264.6 g
The molar volume at STP (22.4 dm3/mol22.4\text{ dm}^3\text{/mol}) converts 67.2 dm367.2\text{ dm}^3 of SO2SO_2 into 3.0 moles3.0\text{ moles}. Accounting for the 90%90\% conversion efficiency yields 2.7 moles2.7\text{ moles} of SO3SO_3. Absorption of SO3SO_3 into concentrated H2SO4H_2SO_4 to form oleum (H2S2O7H_2S_2O_7) followed by dilution with water yields a 1:11:1 molar ratio of H2SO4H_2SO_4 relative to SO3SO_3. Multiplying 2.7 moles2.7\text{ moles} by the molar mass of H2SO4H_2SO_4 (98 g/mol98\text{ g/mol}) gives the correct mass of 264.6 g264.6\text{ g}.

Step-by-Step Solution

1
Calculate the moles of SO2SO_2 gas at STP.
n(SO2)=67.2 dm322.4 dm3/mol=3.0 molesn(SO_2) = \frac{67.2\text{ dm}^3}{22.4\text{ dm}^3\text{/mol}} = 3.0\text{ moles}
Molar volume of any ideal gas at STP is 22.4 dm3/mol22.4\text{ dm}^3\text{/mol}.
2
Apply the 90%90\% conversion efficiency to find the moles of SO3SO_3 produced.
n(SO3)=3.0 mol×0.90=2.7 molesn(SO_3) = 3.0\text{ mol} \times 0.90 = 2.7\text{ moles}
Only 90%90\% of the reacted SO2SO_2 is converted to SO3SO_3 under operating conditions.
3
Relate the moles of SO3SO_3 to the moles of H2SO4H_2SO_4 produced.
n(H2SO4)=n(SO3)=2.7 molesn(H_2SO_4) = n(SO_3) = 2.7\text{ moles}
The absorption of SO3SO_3 into concentrated H2SO4H_2SO_4 forms oleum (H2S2O7H_2S_2O_7), which upon dilution with water yields H2SO4H_2SO_4 with an overall 1:11:1 stoichiometric molar equivalence to SO3SO_3.
4
Calculate the total mass of H2SO4H_2SO_4 formed.
Mass=2.7 mol×98 g/mol=264.6 g\text{Mass} = 2.7\text{ mol} \times 98\text{ g/mol} = 264.6\text{ g}
Mass equals number of moles multiplied by molar mass.

Key Concept

Stoichiometry of the Contact Process including STP gas conversion and oleum dilution stoichiometry
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Topics

Tetraoxosulfate(VI) Acid: Contact Process and Properties

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