According to the balanced chemical equation:
N₂ + 3H₂ → 2NH₃
The stoichiometry of the equation shows that for every 3 moles of hydrogen gas (H₂) that react, 2 moles of ammonia (NH₃) are produced.
Since the volume of ammonia produced is given as 8.80 L at STP, we can use the ideal gas law to find the number of moles of ammonia produced:
PV = nRT
n = PV/RT
At STP (standard temperature and pressure), the values are:
P = 1 atm
V = 8.80 L
T = 273.15 K
R = 0.08206 L·atm/(mol·K)
n = (1 atm x 8.80 L) / (0.08206 L·atm/(mol·K) x 273.15 K) = 0.376 mol
From the balanced equation, we know that 3 moles of hydrogen gas react with 2 moles of ammonia produced. Therefore, the number of moles of hydrogen gas required is:
0.376 mol NH₃ x (3 mol H₂/2 mol NH₃) = 0.564 mol H₂
Finally, we can use the ideal gas law again to calculate the volume of hydrogen gas required at STP:
PV = nRT
V = nRT/P
At STP (as above), the value of n is 0.564 mol. Thus:
V = (0.564 mol x 0.08206 L·atm/(mol·K) x 273.15 K) / 1 atm = 13.2 L
Therefore, 13.2 liters of hydrogen gas are needed to form 8.80 L of NH₃ at STP.