What mass of aluminium oxide is produced when 54.0 g of aluminium reacts with excess oxygen?
4Al + 3O(small 2) ---> 2Al(small 2)O(small 3)
And can you please explain how you did it? I'm just trying to remember myself.
2007-11-26 09:50:46 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
Atomic weights: Al=27 O=16 Al2O3=102
4Al + 3O2 ===> 2Al2O3
54.0gAl x 1molAl/27gAl x 2molAl2O3/4molAl x 102gAl2O3/1molAl2O3 = 102g Al2O3
2007-11-26 10:03:51 · answer #1 · answered by steve_geo1 7 · 0⤊ 0⤋
since the oxygen is in excess, this means that the Al is the limiting reagent.
54.0g Al x (1 mol Al/26.98g Al) x (2mol Al2O3/4mol Al) x (101.96g Al2O3/1mol Al2O3)
What that is is converting Al to moles using it's molar mass, and then using the ratio of moles between Al and Al2O3 to find moles of Al2O3, and then use molar mass of that to find how many moles of it there are.
2007-11-26 17:57:43 · answer #2 · answered by Stephen Z 2 · 0⤊ 0⤋