Answer:
29.9L of NH₃ are produced
Step-by-step explanation:
The reaction of H2 with N2 to produce NH3 is:
3 H2 + N2 → 2 NH3
For a complete reaction of 2.00 moles of H2 there are required:
2.00 moles H2 * (1 mole N2 / 3.00 moles H2) = 0.667 moles N2
The limiting reactant is hydrogen.
The volume of hydrogen using PV = nRT is (STP: 273.15K; 1atm):
V = nRT / P
V = 2.00mol * 0.082atmL/molK*273.15K / 1atm
V = 44.8L
As we know for the reaction, 3 moles of hydrogen produce 2 moles of NH3.
The moles of H2 produced are:
2mol H2* (2mol NH3 / 3mol H2) = 1.33 moles NH3 are produced
Under pressure and temperature constant (STP) you can apply The Avogadro's law that states:
V₁ / n₁ = V₂ / n₂
Where V is volume and n moles of 1, the first gas and 2, the second gas.
V₁ / n₁ = V₂ / n₂
44.8L / 2.00mol = V₂ / 1.33 moles
V₂ =
29.9L of NH₃ are produced