How much sulfite acid can be obtained as a result of the interaction of sulfur oxide (4) with water weighing 72 g?

Let’s write down given:

m (H2O) = 72 g

+ SO 2

m (H2SO3) -?

Decision:

We write the equation for the reaction of the interaction of sulfur (VI) oxide with water, with the formation of sulfurous acid:

SO 2 + H2O = H2SO3

1 mol 1 mol

M (H2O) = 1 * 2 + 16 * 1

M (H2O) = 18 g / mol

M (H2SO3) = 1 * 2 + 32 * 1 + 16 * 3

M (H2SO3) = 82 g / mol

Let’s compose the ratio:

From the reaction equation we see that from 18 g (H2O) 82 g (H2SO3) are formed,

Then from 72 g (H2O) x g (H2SO3) is formed, hence

X = 72g * 82g / 18g

X = 328 g (H2SO3)

Answer: from 72 g of water, 328 g of sulfurous acid is formed.