# chemistry

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Aluminum is burned in O2 to give Al2O3. 74g of aluminum are mixed and reacted with 56g O2. What mass of aluminum oxide is produced?

#gAl2O3 = 56gO2 x ??

• chemistry -

You have BOTH Al ad O2 given; therefore, this is a limiting reagent problem. Convert g Al to moles Al, convert grams O2 to moles O2, determine the limiting reagent, THEN you can determine the amount of Al2O3 formed.

• chemistry -

I got 2.74mol Al and 1.75mol O2, how do I find which one is iin excess? what are the next steps?

• chemistry -

If you had written the balanced equation you probably could have figures out how to do it.
4Al + 3O2 ==> 2Al2O3

So 2.74 moles Al will produce how much Al2O3 (given all of the oxygrn needed)? That will be 2.74 moles Al x (2 mol Al2O3/4 mol Al) = 2.74 x (2/4) = 1.37 mole Al2O3.

How much will the oxygen produce(given all the Al needed)? That will be
1.75 mole O2 x (2 mol Al2O3/3 mol O2) = 1.17 mole Al2O3
Obviously both can't be right. So the combination of the two will produce as much as the SMALlER one; therefore, oxygen is the limiting reagent, you will have 1.17 mole Al2O3 produced, and the mass of the Al2O3 will be 1.17 x molar mass Al2O3. You could calculate, if you wish, how much of the Al is used, subtract from the amount initially, to find the amount remaining unreacted.

• chemistry -

So the grams of Al2O3 is 119.34 :)
thanks :)

• chemistry -

If your prof is picky about the number of significant figures, you have more in your answer than allowed. You had 74 and 56 g respectively, both have two s.f.; therefore, you are allowed 2 in the answer.

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