Answer
$19.4$ g of $O_2$ are needed in this reaction.
Work Step by Step
1. Calculate the molar mass of $Na_2O$:
$Na: 22.99g * 2= 45.98g $
$O: 16.00g$
45.98g + 16.00g = 61.98g
$ \frac{1 mole (Na_2O)}{ 61.98g (Na_2O)}$ and $ \frac{ 61.98g (Na_2O)}{1 mole (Na_2O)}$
2. The balanced reaction is:
$4Na + O_2 --\gt 2Na_2O$
According to the coefficients, the ratio of $Na_2O$ to $O_2$ is 2 to 1:
$ \frac{ 1 mole(O_2)}{ 2 moles (Na_2O)}$ and $ \frac{ 2 moles (Na_2O)}{ 1 mole(O_2)}$
3. Calculate the molar mass for $O_2$:
$O: 16.00g * 2= 32.00g $
$ \frac{1 mole (O_2)}{ 32.00g (O_2)}$ and $ \frac{ 32.00g (O_2)}{1 mole (O_2)}$
4. Use the conversion factors to find the mass of $O_2$
$75.0g(Na_2O) \times \frac{1 mole(Na_2O)}{ 61.98g( Na_2O)} \times \frac{ 1 mole(O_2)}{ 2 moles (Na_2O)} \times \frac{ 32.00 g (O_2)}{ 1 mole (O_2)} = 19.4g (O_2)$
