Write balanced chemical equations to show how $S O_{3}$ is converted to Sulphuric acid at the contact process.

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Solution

$S O_{3}$ is converted to Sulphuric acid in the contact process by dissolution in 98% sulphuric acid to form oleum ( $H_{2} S_{2} O_{7}$ ) followed by dilution of oleum with water.
$S O_{3} + H_{2} S O_{4} \to H_{2} S_{2} O_{7}$
$H_{2} S_{2} O_{7} + H_{2} O \to 2 H_{2} S O_{4}$

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Write balanced chemical equations to show how SO3 is converted to Sulphuric acid at the contact process.

SO3 is converted to Sulphuric acid in the contact process by dissolution in 98% sulphuric acid to form oleum (H2S2O7) followed by dilution of oleum with water.SO3+H2SO4→H2S2O7H2S2O7+H2O→2H2SO4

Write balanced chemical equations to show how $S O_{3}$ is converted to Sulphuric acid at the contact process.

$S O_{3}$ is converted to Sulphuric acid in the contact process by dissolution in 98% sulphuric acid to form oleum ( $H_{2} S_{2} O_{7}$ ) followed by dilution of oleum with water.
$S O_{3} + H_{2} S O_{4} \to H_{2} S_{2} O_{7}$
$H_{2} S_{2} O_{7} + H_{2} O \to 2 H_{2} S O_{4}$