# 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?

$$\ce{3 H2(g) + N2(g) -> 2 NH3(g)}$$

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 did it the hard way, but correctly except let's keep units in the density. You value is off slightly due to rounding of a few terms (with the simplistic assumption that they are ideal gasses). – Joseph Hirsch Dec 12 '16 at 5:01
• Well Laplacian, there is a flaw with this question because you need to specify the temperature. PV=nRT and we already know what R is and the volume but we dont know the temperature nor do we know the pressure. So therefore there's too much assumptions that we make and the liter to mole ratio is skewed, you needed to specify whether it was STP or something else – John Rawls Dec 12 '16 at 7:03
• @JohnRawls Sorry, the question was supposed to say that it was at STP (that's where the density came from). – Teoc Dec 12 '16 at 7:28
• @QuantumAMERICCINO While the question may look like homework to you, it is actually not homework. Please do not randomly tag questions you think are homework as homework. – Teoc Dec 12 '16 at 7:30
• Doesn't have to be STP as long as both gasses are at same temp miles are proportional to volume. – Joseph Hirsch Dec 12 '16 at 8:12

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