Question
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 %
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 %
Answers
It looks right to me. Be sure to note that Al is the limiting reagent!
http://danielleamorim.tripod.com/
http://danielleamorim.tripod.com/
that's is a wrong answer because you must divide the actual yield by the theoretical yield so the answer is 172.41 % and the lab is totally failure
Related Questions
Consider the following reaction:
2N2O(g)--->2N2(g) + O2(g)
a. express the rate of the reaction...
The Cu2+ ions in this experiment are produced by the reaction of 1.0g of copper.
Information alre...
15.0g of potassium trioxocarbonate(v) was crushed as heated with 0.1g of manganese (iv)oxide.
1) wr...
Hydrogen gas can be produced by the reaction of magnesium metal with hydrochloric acid by the reacti...