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How many grams of aluminum oxide are produced according to the reaction below given that you start with 13.2 grams of Al and 13.2 grams of O2?
Reaction: 4AI + 3 02 → 2A1203

Sagot :

Answer: 24.87g Al2O3

Explanation: Aluminum is our limiting reagent

[tex]13.2g Al * \frac{1 mol Al}{26.98 g} * \frac{2 mol Al2O3}{4 mol Al} = 0.244 mol Al2O3[/tex]

First of all we need to convert the grams of aluminium to moles, then use the molar fraction of the balanced equation (4 moles of aluminium equals 2 of aluminum oxide).

If we made the same procedure with the oxigen we get a 0.273 mol of Al2O3, therefore the O2 is the excess reagent.

The last step is convert the moles of the limiting reagent to grams

[tex]0.244 mol Al_{2} O_{3} * \frac{101.96g}{1mol Al2O3} = 24.87g Al2O3[/tex]

And that´s it!

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