Answer:
4.7 g. Option 5 is the right one.
Explanation:
4Al(s) + 3O₂ (g) ⇄ 2Al₂O₃ (s)
We convert the mass of reactants to moles, in order to determine the limiting.
2.5 g Al / 26.98 g/mol → 0.092 moles of Al
2.5 g O₂ / 32g/mol → 0.078 moles of O₂
Ratio is 4:3. 4 moles of Al react with 3 moles of O₂
Then, 0.092 moles of O₂ would react with (0.092 . 3)/ 4 = 0.069 moles O₂
We have 0.078 moles of O₂ and we need 0.069 moles, the oxygen is the limiting in excess. Therefore the Al is the limiting reactant.
Ratio is 4:2. 4 moles of Al, can produce 2 moles of Al₂O₃
Then, 0.092 moles of Al would produce (0.092 .2) / 4 = 0.046 moles
If we convert the moles to mass, we find the anwer:
0.046 mol . 101.96 g/mol = 4.69 g
