Mole ConceptHard
Question
A certain mixture of MnO and MnO2 contains 66.67 mol percent of MnO. What is the approximate mass percent of Mn in it? (Mn = 55)
Options
A.66.67
B.24.02
C.72.05
D.69.62
Solution
Mass per cent of Mn = $\frac{1 \times 55}{\frac{2}{3} \times 71 + \frac{1}{3} \times 87} \times 100 = 72.05\%$
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