Answer
There is a total of 0.146 g of $H_2$ in 1620 mL of $H_2$ gas at STP conditions.
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 :
$H: 1.008g * 2= 2.016g $
$ \frac{1 \space mole \space (H_2)}{ 2.016 \space g \space (H_2)}$ and $ \frac{ 2.016 \space g \space (H_2)}{1 \space mole \space (H_2)}$
$\frac{1 \space L}{1000 \space mL}$ and $\frac{1000 \space mL}{1 \space L}$
2. Calculate the volume of that sample:
$1620 \space mL \space (H_2) \times \frac{1 \space L}{1000 \space mL} \times \frac{1 \space mole \space (O_2)}{22.4 \space L \space (O_2)} \times \frac{ 2.016 \space g \space (H_2)}{1 \space mole \space (H_2)} = 0.146 \space g \space (H_2)$
