How many moles and grams of aluminum can be obtained by reducing it from 126 g of aluminum oxide?

Let’s implement the solution:
1. By the condition of the problem, we write the equation:
m = 126 g. X g. -? X mole -?
Al2O3 + 3H2 = 2Al + 3H2O – OBP, aluminum obtained;
2. Calculation of molar masses:
M (Al2O3) = 101.8 g / mol;
M (Al) = 26.9 g / mol.
3. Find the number of moles:
Y (Al2O3) = m / M = 126 / 101.8 = 1.24 mol.
4. Proportion:
1.24 mol (Al2O3) – X mol (Al);
-1 mol – 2 mol from here, X mol (Al) = 1.24 * 2/1 = 2.48 mol.
5. Determine the mass of the product:
m (Al) = Y * M = 2.48 * 26.9 = 66.7 g.
Answer: the number of moles of aluminum is 2.48, the mass is 66.7 g.