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) --> 2k2o(s)
44%
75%
22%
88%

88%

Explanation:

4 K(s) + O2(g) --> 2K2O(s)

Since Oxygen gas is said to be in excess, it means that potassium is the limiting reactant.

From the reaction above;

4 mol of K produces 2 mol of K2O

In terms of mass;

Mass = Number of moles * Molar mass

156.3752g (4mol * 39.0983g/mol) of K produces 188.3732g (2mol * 94.1866g/mol) of K2O

If 156.3752g produces 110.1856g, 162 would produce?

156.3752 = 188.3732

162 = x

Upon solving for x,

x = (162 * 188.3732) / 156.3752

x =  195.1490g

Theoretical yield =  195.1490g

But Experimental yield =  172 g

Percent yield = (Experimental Yield / Theoretical Yield) * 100

Percent Yield = (172 / 195.1490) * 100

Percent Yield = 0.8820 * 100 = 88%

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