V (O2) = 4.48 L
m (Al2O3) -?
1) We compose the reaction equation corresponding to the condition of the problem:
4Al + 3O2 = 2Al2O3;
2) Find the amount of oxygen contained in 4.48 liters of gas:
n (O2) = V: Vm = 4.48 L: 22.4 L / mol = 0.2 mol;
3) We compose logical equality:
if 2 mol of Al2O3 requires 3 mol of O2,
then x mol of Al2O3 will require 0.2 mol of O2,
then x = 0.2 * 2: 3 = 0.13 mol.
4) Find the mass of aluminum oxide formed during the reaction:
m (Al2O3) = n * M = 0.13 mol * 102 g / mol = 13.26 g.
Answer: m (Al2O3) = 13.26 g.
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