Answer
The volume occupied by 6.40 g of $O_2$ gas is equal to 4.48 L
Work Step by Step
1. Setup the conversion factors:
For a gas at STP conditions, we can use the equality: $1 \space mole = 22.4 \space L$:
$\frac{1 \space mole \space (O_2)}{22.4 \space L \space (O_2)}$ and $\frac{22.4 \space L \space (O_2)}{1 \space mole \space (O_2)}$
Molar mass of $O_2$:
$O: 16.00g * 2= 32.00g $
$ \frac{1 \space mole \space (O_2)}{ 32.00 \space g \space (O_2)}$ and $ \frac{ 32.00 \space g \space (O_2)}{1 \space mole \space (O_2)}$
2. Calculate the volume of that sample:
$6.40 \space g \space (O_2) \times \frac{1 \space mole \space (O_2)}{ 32.00 \space g \space (O_2)} \times \frac{22.4 \space L \space (O_2)}{1 \space mole \space (O_2)} = 4.48 \space L \space (O_2)$
