What mass of sulfuric acid is required to completely dissolve 8.1 g of aluminum?

To solve, we compose the equation:
m = 8.1 g. X g. -?
1.2Al + 3H2SO4 = Al2 (SO4) 3 + 3H2 – OBP, hydrogen, aluminum sulfate is released;
2. Calculations:
M (Al) = 26.9 g / mol;
M (H2SO4) = 98 g / mol.
3. Determine the amount of the original substance:
Y (Al) = m / M = 8.1 / 26.9 = 0.3 mol.
4. Proportion:
0.3 mol (Al) – X mol (H2SO4);
-2 mol -3 mol from here, X mol (H2SO4) = 0.3 * 3/2 = 0.45 mol.
5. Find the mass of the product:
m (H2SO4) = Y * M = 0.45 * 98 = 44.1 g.
Answer: sulfuric acid was obtained with a mass of 44.1 g.