Question

Given the following chemical equation, if 162.0 grams of potassium reacts with excess oxygen gas and was found to produce 172 grams of potassium oxide, what is the percent yield for this reaction? 4 K(s) + O2(g) arrow 2K2O(s)

Answer 1

4 K(s) + O2(g) arrow 2K2O(s)

Answer 2

find mole of K

Answer 3

As potassium is the limiting reagent, the amount of K reacted initially will determine the amount of K20 theoretically yield. As oxygen is in excess, it can only react with whatever K is present, so it will have no effect on the amount of product we form.

The balanced chemical equation above shows us that for every 4 moles of K we have, it will need to react with only 1 mole of oxygen to produce 2 moles of K20.

Divide 162.0 g by the atomic mass of K (found from the periodic table) to find the number of moles of K we have. Then, compare this value with the 4:2, or 2:1 molar ratio between K and K20 to find the number of moles of K20 that could theoretically be formed. Then, multiply this theoretical number of moles by the relative molecular mass of K20 (work out from periodic table again) to find the theoretical number of grams of K20 formed.

Then,
% Yield = (Actual Yield in grams (172 g) / Theoretical Yield in grams) x 100

Answer 4

mole = mass/Mr
mole of K= 162/ (39.1*4)
mole= .....

now u have mole of K
see the ratio between K and K2O
as u see its 4:2
so mole that you found--4
x moles--2
cross multiply and you'll get mole of K2O
one you get it find mass of K2O
mass= mole*Mr
mass= mole that u just found* ((39.1*2+16)*2 ) (its times 2 because coefficient before K2O is 2)
and you'll get mass

now % yield= (given mass /mass that you just found) *100