Question #31898
What is the percent yield for the reaction of 6.92 g of K and 4.28 g of O2 if 7.36 g of KO2 is recovered in the laboratory?
Expert's answer
Potassium superoxide (KO2) is produced by burning molten potassium (K) in pure oxygen (O2) according to following reaction:
K + O2 = KO2
The quantity of potassium is
n(K) = m(K)/M(K) = 6.92 g / 39g/mol = 0.177 mol
The quantity of oxygen is
n(O2) = m(O2)/M(O2) = 4.28 g / 32g/mol = 0.134 mol (The Limiting Reagent)
The quantity of potassium peroxide is
n(KO2) = m(KO2)/M(KO2) = 7.36 g / 71 g/mol = 0.104 mol
If we assume that the yield of the reaction is 100%, the mass of potassium peroxide will be:
m(100% KO2) = 0.134 mol * 71 g/mol = 9.51 g
Thus, theyield of the reaction is:
yield,% = m(KO2) / m(100% KO2) = (7.36 g / 9.51 g ) *100% = 77.39%
K + O2 = KO2
The quantity of potassium is
n(K) = m(K)/M(K) = 6.92 g / 39g/mol = 0.177 mol
The quantity of oxygen is
n(O2) = m(O2)/M(O2) = 4.28 g / 32g/mol = 0.134 mol (The Limiting Reagent)
The quantity of potassium peroxide is
n(KO2) = m(KO2)/M(KO2) = 7.36 g / 71 g/mol = 0.104 mol
If we assume that the yield of the reaction is 100%, the mass of potassium peroxide will be:
m(100% KO2) = 0.134 mol * 71 g/mol = 9.51 g
Thus, theyield of the reaction is:
yield,% = m(KO2) / m(100% KO2) = (7.36 g / 9.51 g ) *100% = 77.39%
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