Question

how many grams of of oxygen are in 135.0 g of potassium permanganate? 158.0 g o 0.2136 g o 54.67 g o 3.417 g o

Explanation:

Step1: Determine the molar mass of \(KMnO_4\)

The molar mass of \(K\) (\(M_{K}\)) is \(39.10\space g/mol\), of \(Mn\) (\(M_{Mn}\)) is \(54.94\space g/mol\), and of \(O\) (\(M_{O}\)) is \(16.00\space g/mol\).

$$M_{KMnO_4}=39.10 + 54.94+4\times16.00=158.04\space g/mol$$

Step2: Calculate the mass fraction of \(O\) in \(KMnO_4\)

There are 4 \(O\) atoms in \(KMnO_4\).
The mass of \(O\) in one mole of \(KMnO_4\) is \(4\times16.00 = 64.00\space g\)
The mass fraction of \(O\), \(w_O=\frac{64.00}{158.04}\)

Step3: Calculate the mass of \(O\) in \(135.0\space g\) of \(KMnO_4\)

$$m_O = 135.0\times\frac{64.00}{158.04}$$

$$m_O=\frac{135.0\times64.00}{158.04}\approx54.67\space g$$

Answer:

\(54.67\space g\space O\) (the third option)