In the haber reaction, patented by german chemist fritz haber in 1908, dinitrogen gas combines with dihydrogen gas to produce gaseous ammonia. this reaction is now the first step taken to make most of the world's fertilizer. suppose a chemical engineer studying a new catalyst for the haber reaction finds that 284. liters per second of dinitrogen are consumed when the reaction is run at 196.°c and 0.75atm . calculate the rate at which ammonia is being produced. give your answer in kilograms per second. be sure your answer has the correct number of significant digits.

Explanation:

1) Word equation (given):

•Dinitrogen gas + dihydrogen gas → gaseous ammonia.

2) Chemical equation:

•N₂ (g) + 3H₂ (g) → 2NH₃ (g)

3) Mole ratios:

•1 mole N₂ (g) : 3 mole H₂ (g) : 2 mole NH₃ (g)

4) Reaction rates:

•Rate of dinitrogen consumption: r₁ = n₁ / t (moles/s)

•Rate of ammonia production: r₂ = n₂ / t (moles/s)

•Due to the stoichiometric ratios: r₂ = 2 × r₁

5) Calculate r₁:

•284 liter / s

•PV = nRT ⇒ n = PV / (RT)

•Divide by time, t: n/t = P (V/t) / (RT)

•Substitute V/t = 284 liter/s, P = 0.75 atm, and T = 196 +273.15K = 469.15K

r₁ = n₁ / t = (0.75 atm) (284 liter/s) / [ (0.08206 atm-liter/K-mol) (469.15k) ]

= 5.53 moles/s

6) Calcualte r₂

7) Convert rate in mole/s to rate in kg/s

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g