What mass of sulfuric acid with a mass fraction of 0.1 is required to dissolve 5.6 grams of iron?

Sulfuric acid with a mass fraction of 0.1 is considered dilute and reacts with iron to form iron (II) sulfate and hydrogen. The reaction equation is as follows:

Fe + H2SO4 = FeSO4 + H2

To solve the problem, we first calculate what mass of pure acid is required for the reaction with a given weight of iron. Then we recalculate the acid with a mass fraction of 0.1.

The molar mass of iron is 56 g / mol

The molar mass of sulfuric acid is 96 g / mol

According to the reaction equation, we make up the proportion:

5.6 g of iron corresponds to X g of sulfuric acid, as

56 g / mol of iron corresponds to 96 g / mol of sulfuric acid

X = (5.6 * 96) / 56 = 9.6 g of pure sulfuric acid reacted.

m (solution H2SO4) = 9.6 / 0.1 = 96 g.

Answer: the mass of the sulfuric acid solution is 96 g.