20.0g sample of al reacts to produce 32.7g of Al2O3. what is the percentage yield of the reaction?

Question

4Al(s) + 3O₂(g) ==> 2Al₂O₃(s) ... balanced equation

use the method

(Al) mole: mass Al: Al3O3 mole: mass (Al2O3)

Answer 1

First start off by writing the equation of what this is going to look like:

4Al+3 + 3 O-2(2) ==> 2 Al2O3

Now that we have a balanced equation we can use stoichiometry to see what will be yielded:

20g of Al

20 g x 1mol/26.9815g grams cancel top to bottom and you have how many moles this is:

0.7412mol of Al

See what this means for moles of Oxygeh:

4mol Al react with 3 mol of O2 according to the equation so:

4 Al ==> 3O2

0.7412 mol ==> X O2

4x = 3(0.7412)/4

x = 0.5559 mol O2

now do your coefficient to substance in moles ratio to see which one limits the reaction by running out first. Which ever number is smaller is the limiting reactant.

0.7412/4 = 0.1853 Al

0.5559/3 =0.1853 O2

So you can proceed the reaction with either. You might as well proceed it with Aluminum since you were given that you have 20g of the substance to react with.

So now

4 mol of Al ==> 2 Al2O3

so for 20g given of Al that is 0.7412 moles so

4 mol of Al ==> 2 Al2O3

0.7412 mol of Al ==>X Al2O3

solve for x

4X =2(0.7412)/4

X = 0.3706 mol Al2O3

now take this to grams of Al2O3

atomic weight of Al2O3 = 101.963 g/mol

now take moles back to grams

0.3706 mol x 101.963g/mol moles will cancel moles top to bottom and you will have grams

37.787g of Al2O3 this is the total amount that can be yielded based on the balanced stoichiometric equation. So if the problem is telling you that only 32.7g came out from the reaction you would need to do a ration to figure out what percent of this actually came out.

so you would take 32.7g/37.787g which is the total that can amount from this reaction and you would get:

32.7/37.787 = 0.865377 x 100 for the percentage and you get 86.5%