2.00 L H2 at 474 torr combines with 1.00 L N2 at 0.200 atm.

Both are at 25 degree C. (A) calculate the partial pressures of each of the gases before reaction to produce ammonia, NH3...N2+3 H2~>2NH3. (B) calculate the moles of each gas before the reaction. (C) determine the limiting reactant. (D) calculate the moles of ammonia produced, and (E) calculate the partial pressures of all species in the system, and (F) the total pressure in the system.

N2 (g) + 3 H2 (g) => 2 NH3 (l)

PV = nRT, where P = Pressure, V = Volume, n = number of moles, R = gas constant, and T = temperature. Rearrange this Ideal-Gas Equation to find n for both H2 and N2:

n = PV / RT
P H2 = (474 torr) / (1 atm / 760 torr) = 0.624 atm
V H2 = 2.00 L
T = 25℃ = 25 + 273 = 298 K
R = 0.0821 L-atm / K-mol

n H2 = (0.624 atm)(2.00 L) / (0.0821 L-atm / K-mol)(298 K)
n H2 = 0.0510 moles

P N2 = 0.200 atm
V N2 = 1.00 L

n N2 = (0.200 atm)(1.00 L) / (0.0821 L-atm / K-mol)(298 K)
n N2 = 0.00817 moles

FOR B.) n H2 = 0.0510 moles, n N2 = 0.00817 moles

A.) Total moles = 0.0592 moles. Partial pressure H2 = n total (RT / V) = (0.0592) [(0.0821)(298) / (2.00)] = 0.724 atm
Partial pressure N2 = 1.45 atm

C.) Limiting Reactant is H2 (g)

D.) Since H2 is the LR, (0.0510 moles H2 / 3) x 2 = 0.0340 moles NH3

E.) PV = nRT
P = nRT / V
P = (0.0340 mol)(0.0821 L-atm / K-mol)(298 K) / (3.00 L) = 0.277 atm
The other partial pressures are in part A.)

F.) P total = P H2 + P N2
P total = (0.724 atm) + (1.45 atm) = 2.17 atm

Hope this helps!