3.534 x 10^3 milligrams of H_2 is equal to how many liters of H_2?

...we could specify standard conditions of temperature and pressure, 1*atm;298 *K, and work from there....which equation do I use here?
$V = \frac{n R T}{P} = \frac{\frac{3.354 \cdot g}{2.019 \cdot g \cdot m o {l}^{-} 1} \times 0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 298 \cdot K}{1 \cdot a t m}$
...I make this about $41 \cdot L$...but you will have to specify your terms.