Calculate the mass of aluminum hydroxide required to react with a solution containing 24.5 g of sulfuric acid.

Let’s write down the solution by composing the reaction equation:
2Al (OH) 3 + 3H2SO4 = Al2 (SO4) 3 + 6H2O – ion exchange reaction, aluminum sulfate was obtained;
Let’s make calculations using the formulas:
M (H2SO4) = 98 g / mol;
M Al2 (SO4) 3 = 341.8 g / mol;
Let’s calculate the number of moles of sulfuric acid, if the mass is known:
Y (H2SO4) = m / M = 24.5 / 98 = 0.25 mol;
Let’s make the proportion:
0.25 mol (H2SO4) – X mol Al2 (SO4) 3;
3 mol                       -1 mol from here, X mol of Al2 (SO4) 3 = 0.25 * 1/3 = 0.08 mol;
We calculate the mass of aluminum sulfate by the formula:
m Al2 (SO4) 3 = Y * M = 0.08 * 341.8 = 27.34 g.
Answer: in the course of the reaction, an aluminum sulfate salt weighing 27.34 g was obtained.