Question

At STP, 16L of N2 and 48L of H2 are mixed. Assume all the reactants are consumed, how many L of NH3 could be produced? N2(g) +3H2(g) --> 2NH3(g)

Answer 1

To determine how many liters of NH3 could be produced, we need to use the stoichiometry of the balanced equation.

According to the balanced equation: N2(g) + 3H2(g) -> 2NH3(g)

For every 1 mole of N2, 2 moles of NH3 are produced. Therefore, the ratio of N2 to NH3 is 1:2.

We'll start by converting the given volumes of N2 and H2 to moles using the ideal gas law:

PV = nRT

For N2:

n = PV/RT = (1 atm)(16 L)/(0.0821 L*atm/mol*K)(273 K) ≈ 0.6 moles

For H2:

n = PV/RT = (1 atm)(48 L)/(0.0821 L*atm/mol*K)(273 K) ≈ 2.4 moles

From the balance equation, we can see that the amount of NH3 produced is directly proportional to the amount of N2(g) consumed. Therefore, using the mole ratio, we can determine the amount of NH3 produced:

NH3 produced = 2 * (0.6 moles) = 1.2 moles

Finally, we can convert moles of NH3 back to liters at STP:

PV = nRT

V = (nRT)/P = (1.2 moles)(0.0821 L*atm/mol*K)(273 K)/(1 atm) ≈ 27.9 L

Therefore, approximately 27.9 liters of NH3 could be produced.