How many liters of hydrogen, H2, are needed to react with 10 liters of nitrogen gas in the reaction forming ammonia? How many liters of hydrogen, H2, are needed to react with 10 liters of nitrogen gas in the reaction forming ammonia?

$$\ce{3 H2(g) + N2(g) -&gt; 2 NH3(g)}$$

My try:: Because we have 10 L of nitrogen gas, we have 10/22.4 moles = 0.446 moles of nitrogen gas, and thus need 0.446 * 3 = 1.338 moles of hydrogen gas. As such, we need 2.676 grams of hydrogen gas, or 2.676/0.09 = 29.8 L of hydrogen gas. However this is not correct. Can someone find my problem?

Note: This reaction occurs at STP.
You are on the right track. Dimensional Analysis is your friend:

$$\frac{10\,\text{L  $\ce{N2}$}}{1} \cdot \frac{1\,\text{mol  $\ce{N2}$}}{22.4 \,\text{L  $\ce{N2}$}}\cdot \frac{3\,\text{mol  $\ce{H2}$}}{1\,\text {mol  $\ce{N2}$}}\cdot \frac{22.4\,\text{L  $\ce{H2}$}}{1\,\text {mol  $\ce{H2}$}}=\,?\,\, \text{L $\ce{H2}$}$$

If you do this, you will realize that the quantities cancel exactly and give you a nice clean number.