What amount and mass of aluminum is required to obtain 10.2gr. aluminum oxide?

Reaction: 4Al + 3O2 = 2Al2O3
The molar mass of aluminum oxide is M = 27 * 2 + 16 * 3 = 102 g / mol (determined according to the periodic table).
The amount of oxide substance n = m / M = 10.2 g / 102 g / mol = 0.1 mol. According to the reaction from 4 mol of aluminum, we get 2 mol of oxide, n (Al) / 4 = 0.1 / 2. n (Al) = 4 * 0.1 / 2 = 0.2 mol. Aluminum mass m = n * M = 0.2 mol * 27 g / mol = 5.4 g.
Answer: 0.2 mol or 5.4 g of aluminum is needed.