Consider the balanced chemical equation for the reaction between nitrogen gas and hydrogen gas to produce ammonia.

Question

A quantity of 24 liters of nitrogen gas, at 25oC and 760 torr, is used in an experiment. What volume of hydrogen gas, at the same temperature and pressure, is needed to react completely with the nitrogen?

N2(g) + 3H2(g) --> 2NH3(g)

Answer 1

N₂(g) + 3H₂(g) ==> 2NH₃(g) ... balanced equation

Since the temperature and pressure do not vary, we can look at this equation and read it as ....

1 L N₂ gas reacts with 3 L H₂ gas to produce 2 L NH₃ gas

Using these relationships, we can set up an equation to find out the volume of H₂ gas that will react with 24 L of N₂ gas:

24 L N₂ gas x 3 L H₂ gas / 1 L N₂ gas = 72 L H₂ gas needed

***********************************************************************************************************************

If you want to use the ideal gas law, and go through a more extensive proof, you would proceed as follows:

PV = nRT

Solve for mols of N₂ (n) when P = 760 mm x 1 atm/760 mm = 1 atm and T =25 + 273 = 298K

n = PV/RT = (1 atm)(24 L) / (0.0821 Latm/Kmol)(298K) = 0.981 mols N2

Moles H₂ = 3 x 0.981 according to the balanced equation

Moles H₂ = 2.943 mols H₂

Converting this to liters, we have...

PV = nRT

V = vRT/P = (2.943)(0.0821)(298) / 1

V = 72.0 L H2 gas needed (same as above answer)