combines with oxygen with an atomic weight of 16, to form the compound aluminum oxide, how much oxygen would be required to react completely with 54g of aluminum?
Well, Al2O3 is the formula for aluminum oxide So you have 2 moles of Al, so you need 3 moles of O which is 1.5 moles of O2 So the answer is 48g
aluminum oxide is Al2O3 , or 2 part Al to 3 part O. 27 grams Al2 *2 / 54 grams Al available = 16 grams of O * 3 / X grams of O solve for X: X = 48 grams of oxygen needed next guy is right....the formula for aluminum oxide is Al2O3..sorry...I readjusted the formula above to show the change
