Consider the following reaction: N₂(g) + 3 H₂(g) → 2 NH₃(g)
What volume of NH₃(g) can be produced from 200.0 L of H₂(g) if the gases are measured at 350°C and 400 atm pressure?

Question

Consider the following reaction: N₂(g) + 3 H₂(g) → 2 NH₃(g) What volume of NH₃(g) can be produced from 200.0 L of H₂(g) if the gases are measured at 350°C and 400 atm pressure?

Quizplus · Accepted Answer

We need to use the ideal gas law equation to solve the problem: PV = nRT First, we need to calculate the number of moles of H₂(g) present in 200.0 L at the given conditions: n = PV/RT = (400 atm * 200.0 L) / (0.08206 L*atm/mol*K * 623 K) = 16.63 mol According to the balanced equation, the reaction produces 2 mol of NH₃(g) for every 3 mol of H₂(g) consumed. Therefore, the number of moles of NH₃(g) produced is: n(NH₃) = (2/3) * n(H₂) = (2/3) * 16.63 mol = 11.09 mol Finally, we can use the ideal gas law again to calculate the volume of NH₃(g) produced at the same conditions: V(NH₃) = n(NH₃) * RT/P = 11.09 mol * 0.08206 L*atm/mol*K * 623 K / 400 atm = 133.3 L

Consider the Following Reaction: N2(g) + 3 H2(g) → 2

Consider the Following Reaction: N₂(g) + 3 H₂(g) → 2