In a reaction between Al and O, how many grams of aluminum are required to react with 50.10L of oxygen?
2Al + 3O2→ 2Al2O3 m = 2*27*50.1/(3*22.4) =40.25 g where 2 mol is amount of aluminium which took part in the reaction according to the chemical equation; 27 g/mol is the molar mass of aluminium; 50.10L is volume of oxygen; 3 mol is amount of oxygen which took part in the reaction according to the chemical equation; 22.4 mol/L is the molar volume of the ideal gas.
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