How many moles of hydrogen will be required for reactions with aluminum oxide if 44.8 liters of water are formed?

Let’s write down the reaction equation and arrange the coefficients:
3H2 + Al2O3 = 2 Al + 3H2O
Now we will find the amount of substance for H2O.
44.8 L = 44.8 * 10 ^ 3 ml. Because density of water 1g / ml. , then the mass and water = 44.8 * 10 ^ 3 grams. n = 44.8 * 10 ^ 3/18 = 2489 mol. Since, according to the equation, the amounts of hydrogen and water are equal, then 2489 mol of hydrogen is required