Answer:
229%
Explanation:
The equation of the reaction is;
4Al(s) + 3O2(g) ----> 2Al2O3(s)
We must first determine the limiting reactant;
Number of moles of Al2O3 produced = mass/molar mass = 72.4g/101.96 g/mol = 0.71 moles
For Al
Number of moles reacted = mass/molar mass = 49.2g/27 g/mol = 1.8 moles
If 4 moles of Al yields 0.71 moles of Al2O3
1.8 moles of Al will yield 1.8 × 0.71/4 = 0.32 moles of Al2O3
For O2
Number of moles reacted = mass/molar mass = 42.6g/32g/mol = 1.33 moles
If 3 moles of O2 yields 0.71 moles of Al2O3
1.33 moles of O2 will yield 1.33 × 0.71/3 = 0.31 moles of Al2O3
Oxygen is the limiting reactant.
% yield = actual yield/ theoretical yield × 100/1
% yield = 0.71 moles/0.31 moles × 100
% yield = 229%