Calculate the mass of the salt formed by the interaction of iron hydroxide with sulfuric acid in an amount of 4.5 mol.

Let’s write the equation, arrange the coefficients according to the condition of the problem:
2Fe (OH) 3 + 3H2SO4 = Fe2 (SO4) 3 + 6H2O – ion exchange reaction, obtained iron sulfate and water;
Let’s make calculations using the formulas:
M (H2SO4) = 98 g / mol;
M Fe2 (SO4) 3 = 399.6 g / mol;
We calculate the amount of moles of iron sulfate, making up the proportion:
4.5 mol (H2SO4) – X mol Fe2 (SO4) 3;
3 mol – 1 mol from here, X mol Fe2 (SO4) 3 = 4.5 * 1/3 = 1.5 mol;
Calculate the mass of ferrous sulfate using the formula:
m Fe2 (SO4) 3 = Y * M = 1.5 * 399.4 = 59.94 g.
Answer: iron sulfate weighing 59.94 g was obtained during the reaction.