Question

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

In the Contact Process for the manufacture of tetraoxosulfate(VI) acid, 44.8 dm344.8\text{ dm}^3 of sulfur(IV) oxide (SO2SO_2) gas measured at standard temperature and pressure (STP) is reacted with excess oxygen over a vanadium(V) oxide (V2O5V_2O_5) catalyst. If the catalytic conversion efficiency of SO2SO_2 to sulfur(VI) oxide (SO3SO_3) is 85%85\%, and all produced SO3SO_3 is subsequently absorbed in concentrated H2SO4H_2SO_4 and hydrated to form pure H2SO4H_2SO_4, what mass of pure H2SO4H_2SO_4 in grams is produced? (Molar mass of H2SO4=98 g mol1H_2SO_4 = 98\text{ g mol}^{-1}, Molar volume of gas at STP =22.4 dm3 mol1= 22.4\text{ dm}^3\text{ mol}^{-1})

Answer: 166.6 g

Answer

The mass of pure H2SO4 produced is 166.6 g.
The correct answer of 166.6 g is derived by converting 44.8 dm³ of SO2 at STP to 2.0 moles, taking 85% of that value to find the actual 1.70 moles of SO3 produced, and multiplying by the molar mass of H2SO4 (98 g/mol).

Step-by-Step Solution

1
Calculate the moles of SO2 gas supplied at STP
Moles of SO2 = 2.0 mol
Dividing the gas volume at STP (44.8 dm³) by the molar gas volume (22.4 dm³/mol) gives the molar quantity.
2
Determine theoretical yield of SO3
Theoretical moles of SO3 = 2.0 mol
From the stoichiometric mole ratio in 2SO2 + O2 -> 2SO3, 2 moles of SO2 produce 2 moles of SO3.
3
Calculate actual moles of SO3 produced considering catalytic efficiency
Actual moles of SO3 = 1.70 mol
Multiplying the theoretical yield (2.0 mol) by the 85% conversion efficiency gives the actual yield of 1.70 mol.
4
Calculate mass of H2SO4 produced from the actual SO3 formed
Mass of H2SO4 = 166.6 g
Overall absorption and hydration converts SO3 to H2SO4 in a 1:1 mole ratio (SO3 + H2O -> H2SO4). Multiplying 1.70 mol by molar mass 98 g/mol yields 166.6 g.

Key Concept

Stoichiometry of the Contact Process involving molar volume at STP and percentage conversion efficiency
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