Posted by sh on Sunday, March 8, 2009 at 7:36pm.
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.
I got 2.74mol Al and 1.75mol O2, how do I find which one is iin excess? what are the next steps?
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.
So the grams of Al2O3 is 119.34 :)
thanks :)
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.