What volume of gas will be released when 196 g of dilute sulfuric acid interacts with 5.4 g of aluminum?

To solve the problem, it is necessary to draw up an equation, arrange the coefficients:
2Al + 3H2SO4 (p-p) = Al2 (SO4) 3 + 3H2 – ОВР;
Calculations of molar masses and number of moles:
M (Al) = 26.9 g / mol; M (H2SO4) = 98 g / mol;
Y (Al) = m / M = 5.4 / 26.9 = 0.2 mol (deficient substance);
Y (H2SO4) = m / M = 196/98 = 2 mol (substance in excess);
Further calculations are made for the substance in deficiency.
Let’s make the proportion:
2 mol (Al) – X mol (H2);
-2 mol -3 mol from here, X mol (H2) = 2 * 3/2 = 3 mol.
Determine the volume of hydrogen:
V (H2) = 3 * 22.4 = 67.2 liters.
Answer: during the reaction, 67.2 liters of hydrogen are released.