In the industrial Contact Process, sulfur(VI) oxide (SO_3) is hydrated to form tetraoxosul… — JAMB UTME

In the industrial Contact Process, sulfur(VI) oxide ( $SO_3$ ) is hydrated to form tetraoxosulfate(VI) acid ( $H_2SO_4$ ). What volume of $SO_3$ gas, measured in $\text{dm}^3$ at s.t.p., is theoretically required to produce $196\text{ g}$ of pure $H_2SO_4$ ? [Molar volume of gas at s.t.p. = $22.4\text{ dm}^3\text{mol}^{-1}$ , $H = 1.0$ , $S = 32.0$ , $O = 16.0$ ]

Answer: 44.8 dm³

Answer

44.8 dm³

The molar mass of

H_2SO_4

98.0\text{ g mol}^{-1}

, so

196\text{ g}

corresponds to

2.0\text{ moles}

. According to the equation

SO_3 + H_2O \rightarrow H_2SO_4

1\text{ mole}

SO_3

yields

1\text{ mole}

H_2SO_4

. Therefore,

2.0\text{ moles}

SO_3

gas is needed, which occupies

2.0 \times 22.4\text{ dm}^3 = 44.8\text{ dm}^3

at s.t.p.

Step-by-Step Solution

Calculate the molar mass of tetraoxosulfate(VI) acid (

H_2SO_4

\text{Molar mass of } H_2SO_4 = 2(1.0) + 32.0 + 4(16.0) = 98.0\text{ g mol}^{-1}

Essential to convert mass of acid to chemical amount in moles.

Calculate the number of moles of

H_2SO_4

196\text{ g}

\text{Moles of } H_2SO_4 = \frac{196\text{ g}}{98.0\text{ g mol}^{-1}} = 2.0\text{ mol}

Finds the quantitative molar requirement.

Calculate the required volume of

SO_3

gas at s.t.p.

\text{Volume of } SO_3 = 2.0\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 44.8\text{ dm}^3

Based on the 1:1 mole ratio of

SO_3

H_2SO_4

and standard molar gas volume.

Key Concept

Molar gas volume and stoichiometric relationships in the Contact Process

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