What volume of gas will be released when 200 g of 49% sulfuric acid solution interacts with sodium sulfite?

The interaction of sodium sulfite with sulfuric acid is equimolar, how many moles of salt react, so many moles of sulfur dioxide will be released:

Na2SO3 + H2SO4 = SO2 + H2O + Na2SO4.

Let’s calculate the amount of substance using the value of molar masses M and the data of the problem.

M H2SO4 is 98 g, which means that it entered into a reaction: ((200 * 49%) / 100%) / 98 = 1 mol.

The volume of the evolved SO2 gas is equal to:

22.4 l / mol * 1 mol = 22.4 l.

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