Answer to Question #58937 in General Chemistry for Nitasha

Question

In a reaction between Al and O, how many grams of aluminum are required to react with 50.10L of oxygen?

Answer 1

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.