Upon dissolution of aluminum in sulfuric acid, a gas was formed, the volume of which is 18.8 dm3.

Upon dissolution of aluminum in sulfuric acid, a gas was formed, the volume of which is 18.8 dm3. Find the mass of aluminum that has reacted.

Let’s execute the solution:
1. Let us write down the equation and arrange the coefficients:
2Al + 3H2SO4 = Al2 (SO4) 3 + 3H2 – OBP, hydrogen, aluminum sulfate is evolved;
2. Calculation of the molar masses of substances:
M (Al) = 26.98 g / mol;
M (H2) = 2 g / mol;
V (H2) = 18.8 dm3 = 18.8 l, since 1 dm3 = 1 l.
3. Let’s make the proportion:
1 mole of gas at normal level – 22.4 liters;
X mol (H2) – 18.8 liters. hence, X mol (H2) = 1 * 18.8 / 22.4 = 0.839 mol.
4. Determine the number of moles of aluminum:
X mol (Al) – 0.839 mol (H2);
– 2 mol – 3 mol from here, X mol (Al) = 2 * 0.839 / 3 = 0.539 mol;
5. Find the mass of the metal: Al;
m (Al) = Y * M = 0.559 * 26.98 = 15.08 g.
Answer: to carry out the process, aluminum with a mass of 15.08 g is required.