At STP how many liters of NH3 can be produced from the reaction of 2.00 mol of N2 with 2.00 mol of H2

29.9L of NH₃ are produced

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 =

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