How many grams of aluminum do you need to take to get 175 grams of manganese from its oxide?

In accordance with the condition of the problem, we write down the equation, select the coefficients:
3MnO2 + 4Al = 3Mn + 2Al2O3 – redox reaction, manganese was obtained;
Let’s make calculations with the formulas:
M (Al) = 28.9 g / mol;
M (Mn) = 54.9 g / mol;
Let us determine the number of moles of manganese, if its mass is known:
Y (Mn) = m / M = 175 / 54.9 = 3.18 mol.
Let’s make the proportion:
3.18 mol (Mn) – X mol (Al);
-3 mol -4 mol from here, X mol (Al) = 3.18 * 4/3 = 4.24 mol.
We find the mass of aluminum by the formula:
m (Al) = Y * M = 4.24 * 28.9 = 122.53 g.
Answer: the mass of aluminum is 122.53 g.