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 ??
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.
To find the mass of aluminum oxide produced, we need to use the stoichiometry of the reaction and the concept of molar mass.
First, let's write the balanced chemical equation for the reaction:
4 Al + 3 O2 → 2 Al2O3
From the equation, we can see that 4 moles of aluminum react with 3 moles of oxygen gas to produce 2 moles of aluminum oxide.
Now, we need to calculate the moles of aluminum and oxygen using their masses and molar masses:
Molar mass of Al = 26.98 g/mol
Molar mass of O2 = 32.00 g/mol
To find the moles of aluminum:
Moles of Al = Mass of Al / Molar mass of Al
Moles of Al = 74 g / 26.98 g/mol
Moles of Al = 2.74 mol
To find the moles of oxygen:
Moles of O2 = Mass of O2 / Molar mass of O2
Moles of O2 = 56 g / 32.00 g/mol
Moles of O2 = 1.75 mol
Next, we look at the stoichiometric ratio from the balanced equation. It tells us that 4 moles of aluminum produce 2 moles of aluminum oxide. Therefore, the ratio of moles is 4:2 or 2:1.
Since the ratio is 2:1, the mole ratio can be applied to find the number of moles of Al2O3 formed. In this case, the moles of Al2O3 should be equal to the moles of aluminum because they react in a 1:1 ratio.
Moles of Al2O3 = Moles of Al
Moles of Al2O3 = 2.74 mol
Finally, to find the mass of aluminum oxide:
Mass of Al2O3 = Moles of Al2O3 x Molar mass of Al2O3
Mass of Al2O3 = 2.74 mol x (2 x molar mass of Al) + (3 x molar mass of O)
Mass of Al2O3 = 2.74 mol x [(2 x 26.98 g/mol) + (3 x 16.00 g/mol)]
Mass of Al2O3 = 2.74 mol x (53.96 g/mol + 48.00 g/mol)
Mass of Al2O3 = 2.74 mol x 101.96 g/mol
Mass of Al2O3 = 279.42 g
Therefore, the mass of aluminum oxide produced is 279.42 grams.