Answer:
18.89 g of Al₂O₃.
Explanation:
The balanced equation for the reaction is given below:
4Al + 3O₂ —> 2Al₂O₃
Next, we shall determine the masses of Al and O₂ that reacted and the mass of Al₂O₃ produced from the balanced equation. This can be obtained as follow:
Molar mass of Al = 27 g/mol
Mass of Al from the balanced equation = 4 × 27 = 108 g
Molar mass of O₂ = 16 × 2 = 32 g/mol
Mass of O₂ from the balanced = 3 × 32
= 96 g
Molar mass of Al₂O₃ = (27×2) + (16×3)
= 54 + 48
= 102 g/mol
Mass of Al₂O₃ from the balanced equation = 2 × 102 = 204 g
SUMMARY:
From the balanced equation above,
108 g of Al reacted with 96 g of O₂ to produce 204 g of Al₂O₃.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
108 g of Al reacted with 96 g of O₂.
Therefore, 10 g of Al will react with
= (10 × 96)/108 = 8.89 g of O₂.
From the calculation made above, we can see clearly that only 8.89 g out of 19 g of O₂ given, reacted completely with 10 g of Al.
Therefore, Al is the limiting reactant and O₂ is the excess reactant.
Finally, we shall determine the mass of aluminium oxide, Al₂O₃, produced from the reaction.
NOTE: in this case, the limiting reactant will be used because it will give the maximum mass of the aluminium oxide, Al₂O₃ as all of it is consumed in the reaction.
Al is the limiting reactant and the mass of aluminium oxide, Al₂O₃ produced can be obtained as follow:
From the balanced equation above,
108 g of Al reacted to produce 204 g of Al₂O₃.
Therefore, 10 g of Al will react to produce = (10 × 204)/108 = 18.89 g of Al₂O₃.
Thus, 18.89 g of Al₂O₃ were obtained from the reaction.
