how many grams of aluminum oxide could you produce using this equation?

Question

if you were given 23.6g of aluminum and 17.9g of oxygen gas how many grams of aluminum oxide could you theoretically produce?

Answer 1

Hi Abby B.,

Steps required:

1) Write a balanced equation for the oxidation (has only integers for coefficients).

2) Divide the given reactants by the molar mass of each species.

3) Compare the ratio of the two values you thus obtain, to the coefficient ratio in the balanced equation. What you are doing here is seeing "how many recipe portions" you have of each ingredient, where one "recipe portion" is the number of moles of ingredient as stated in the balanced equation. So for example, if you had 32 g of O₂ (with molar mass ~32) and the equation had "1 O₂ " in it (which it does not!), you would have 1 "recipe portion" of that ingredient.

4) If the recipe portions are not equal, then one ingredient or the other is in excess.

5) Use the "recipe portions" of the not in excess ingredient, to "work the reaction through". So for example, if you had 0.2 recipe units of oxygen, and that was the not in excess ingredient, you would ratio that through the reaction to product. Since the balanced reaction has 3 as coefficient for O₂ and 2 as coefficient for Al₂O₃ , you would obtain 0.2 * (2/3) ~0.133 "recipe portions" of the product Al₂O₃ .

6) Convert the "recipe portions" of Al₂O₃ into moles of Al₂O₃ (multiply by 2).

7) Convert that number of moles of Al₂O₃ into grams, using the molar mass of Al₂O₃ .

You should note that different textbooks may work the mechanics of steps 3-6 differently, to arrive at the same result. I've shown you the way that works for me, not just because I am a chemist, but because I am a cook, and have to scale up recipes frequently! And sometimes ingredients are limited, and you can "short" them -- but you can't do that in chemistry.

-- Cheers, --Mr d.