Answer:
The reactant in excess is Al (s)
The mass of remaining excess reactant is 8.76 g.
Explanation:
This is the reaction:
4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s)
4 moles of solid aluminun reacts with 3 moles of oxygen, to make 2 moles of alumina.
Molar mass Al: 26.98 g/m
Molar mass O₂ : 32.00 g/m
Mass / Molar mass : Moles
20 g Al / 26.98 g/m = 0.741 moles
10 g O₂ / 32.00 g/m = 0.312 moles
Let's find out the limiting reactant and the reactant in excess by rules of three:
4 moles of Al reacts with 3 moles of O₂
0.741 moles of Al reacts with (0.741 .3)/4 = 0.555 moles
Now we know that are needed 0.555 moles of O₂ to consume all the Al but we only have 0.312 moles of O₂.
The O₂ is the limiting reactant, so the Al is the reactant in excess.
Let's make another rule of three
3 moles of O₂ reacts with 4 moles of Al
0.312 of O₂ reacts with (0.312 .4)/3 = 0.416
We need 0.416 moles to consume all the O₂ and we have 0.741 moles of Al. There's still Al to consume, that's why the Al is the reactant in excess.
When we use 0.416 moles of 0.741 moles of Al, 0.325 moles are remaining to consume all the Al.
Let's calculate the mass
0.325 moles . 26.98 g/m = 8.76 grams
