If the reaction of 2.5 g of Al with 2.5 g of O2 produced 3.5 g of Al2O3, what is the percent yield of Al2)3? Be sure to write and balance equation.
4Al + 302 ---> 2Al2O3
2.5 g Al x (1 mol Al / 108 g) x (2 mol Al2O3 / 4 mol Al) = .0115 mol Al2O3
.0115 mol Al2O3 (204 g Al2O3 / 1 mol Al2O3) = 2.36 Al2O3
Percent Yield = (2.36 / 3.50) X 100 = 67 %
Well, if my calculations are correct (and I promise they are, because clowns are known for their math skills), the percent yield of Al2O3 is 67%. That means we're 67% successful in creating the Al2O3, and the other 33% must have been lost somewhere. Maybe it went on vacation or decided to join the circus - who knows? But hey, 67% is not too shabby!
To find the percent yield of Al2O3 in this reaction, we first need to balance the equation:
4Al + 3O2 ---> 2Al2O3
Next, we need to calculate the number of moles of Al2O3 produced from the given mass of Al:
Mass of Al = 2.5 g
Molar mass of Al = 27 g/mol
Number of moles of Al = Mass of Al / Molar mass of Al
= 2.5 g / 27 g/mol
= 0.093 mol Al
Using the stoichiometry of the balanced equation, we can determine the number of moles of Al2O3 produced:
1 mol Al2O3 = 2 mol Al
X mol Al2O3 = 0.093 mol Al
Therefore, X = 0.093 mol Al2O3
Now, we can calculate the mass of Al2O3 produced:
Molar mass of Al2O3 = 102 g/mol
Mass of Al2O3 = Number of moles of Al2O3 x Molar mass of Al2O3
= 0.093 mol Al2O3 x 102 g/mol
= 9.486 g Al2O3
Finally, we can calculate the percent yield using the actual yield (mass of Al2O3 produced) and the theoretical yield (mass of Al2O3 that would be produced if the reaction went to completion):
Percent Yield = (Actual yield / Theoretical yield) x 100
= (9.486 g / 3.5 g) x 100
= 270.17 %
Therefore, the percent yield of Al2O3 in this reaction is 270.17%.
It looks right to me. Be sure to note that Al is the limiting reagent!
http://danielleamorim.tripod.com/