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ammonia, NH3 is an important industrial chemical. it is produced using the haber process from nitrogen, N2 and hydrogen, H2. how much ammonia can be produced from 28 kg of nitrogen and 2 kg of hydrogen, assuming the reaction yield is 90%

User Votive
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2 Answers

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Final answer:

The Haber process is used to produce ammonia (NH3) from nitrogen (N2) and hydrogen (H2). Given the mass of nitrogen and hydrogen, we can calculate the amount of ammonia produced using the reaction stoichiometry. With a reaction yield of 90%, approximately 900 moles of ammonia can be produced.

Step-by-step explanation:

The Haber process is used to produce ammonia (NH3) from nitrogen (N2) and hydrogen (H2). The balanced chemical equation for this reaction is:

N2 + 3H2 → 2NH3

To determine the amount of ammonia that can be produced, we need to calculate the limiting reactant and use the reaction stoichiometry. Let's start by converting the given mass of nitrogen and hydrogen into moles:

  1. Mass of nitrogen = 28 kg × (1000 g/kg) ÷ (28 g/mol) = 1000 mol
  2. Mass of hydrogen = 2 kg × (1000 g/kg) ÷ (2 g/mol) = 1000 mol

Since the reaction stoichiometry is 1:3 for nitrogen to hydrogen, we have equal moles of both reactants, resulting in 1000 mol of ammonia being produced.

Given that the reaction yield is 90%, we multiply the calculated amount of ammonia by the yield:

Ammonia produced = 1000 mol × 0.90 = 900 mol

User Natalie Perret
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Answer:

mass of ammonia = 10.090 kg

Step-by-step explanation:

We have the following chemical reaction:

N₂ + 3 H₂ → 2 NH₃

number of moles = mass / molecular weight

number of moles of N₂ = 28 / 14 = 2 kmoles

number of moles of H₂ = 2 / 2 = 1 kmole

We see that the limiting reactant is hydrogen (H₂).

Now we devise the following reasoning:

if 3 kmoles of hydrogen produces 2 kmoles of ammonia

then 1 kmole of hydrogen produces X kmoles of ammonia

X = (1 × 2) / 3 = 0.66 kmoles of ammonia

But the yield is 90% so the real quantity of ammonia produced is:

(90/100) × 0.66 = 0.594 kmoles of ammonia

mass = number of moles × molecular weight

mass of ammonia = 0.594 × 17 = 10.090 kg

User John Batdorf
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