Calculate the mass of aluminium oxide which contains double the number of oxygen atoms in 192 g of oxygen gas.
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1. Number of moles in 192 g of oxygen gas = $$\begin{gathered} \frac{given mass}{molar mass}=\frac{192}{32}=6 moles. \\ Number of oxygen atoms in 6 moles = 2 \times 6\times N_A (as oxygen gas is O_2) \\ = 12 N_A \\ 2. Number of moles of oxygen atom = \frac{number ofatoms}{N_A}=\frac{12N_A}{N_A}=12 moles \\ Mass of 32 moles of oxygen atom= 12\times 16 = 192 g \end{gathered}$$

3. In 1 mole of Al₂O₃, 54 g of Al is present with 48 g of oxygen.
 
  thus 1 g of oxygen is present with $\frac{54}{48}$g of aluminium
and 512 g of oxygen is present with $\frac{54}{48}\times 192=216 g of aluminium$

Total mass of aluminium oxide =192+216  = 408 g.