Se hacen reaccionar 100 g del agente oxidante con 25 g del agente reductor, según la reacción REDOX LaTeX: N_2\left(g\right)+H_2\left(g\right) \longrightarrow NH_3\left(g\right)N 2 ( g ) + H 2 ( g ) ⟶ N H 3 ( g ) Indique el reactivo en exceso y los gramos de amoniaco formado, si la eficiencia del proceso es del 80 %

Excess reactant: H₂

Mass of produced ammonia,  97.1 g

Explanation:

Identify the reaction:

N₂ + 3H₂  → 2NH₃

We identify the reducing agent and the  oxidizing agent:

N₂ changed the oxidation state from 0 to -3. This is the reduction, so it is the  oxidizing agent. By the way the H₂ is the reducing agent.

We convert the mass to moles:

100 g / 28 g/mol = 3.57 moles of N₂

25 g / 2 g/mol = 12.5 moles of H₂

Ratio is 1:3. For 1 mol of nitrogen, we need 3 moles of hydrogen

Then, 3.57 moles of N₂ would need (3.57 . 3) / 1 = 10.7 moles

We have 12.5 moles of H₂, so the hydrogen is the excess reactant and the nitrogen is the limiting.

To produce ammonia, the reaction needs 1 mol of N₂, that can produce 2 moles of product

3.57 moles of N₂ will produce (3.57 . 2) / 1 = 7.14 moles of NH₃

As yield reaction is 80%, we will produce 7.14 mol . 0.80 = 5.71 moles

We convert the moles to mass: 5.71 mol . 17 g / 1mol = 97.1 g

it is the most commonly accepted one

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