Respuesta :

romannp

here is the answer:

if we take the molecular formula, then

28 g Nitrogen forms 17 g ammoni.

so, 11 g Nitrogen forms 17/28 × 11=6.67 g ammonia.

Hence 6.67 g NH3 is formed by 11 g N2

CintiaSalazar

Answer:

13.35 grams of NH₃ is made if 11 grams of N₂ are used.

Explanation:

First of all you should know that the balanced reaction is:

N₂(g) + 3 H₂-> 2 NH₃(g)

Then, it is possible to determine the mass of each compound that reacts or is produced in the reaction by knowing the atomic mass of each element:

  • N: 14 g/mole
  • H: 1 g/mole

Then:

  • N₂: 2*14 g/mole= 28 g/mole
  • H₂: 2*1 g/mole= 2 g/mole
  • NH₃: 14 g/mole+3*1 g/mole= 17 g/mole

By reaction stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) you can see that they react or are obtained in moles:

  • N₂: 1 mole
  • H₂: 3 moles
  • NH₃: 2 moles

Then, by the reaction stoichiometry of the reaction, you can see the amounts of mass that react or are obtained:

  • N₂: 1 mole*28 g/mole= 28 g
  • H₂: 3 moles*2 g/mole= 6 g
  • NH₃: 2 moles*17 g/mole=  34 g

To calculate the amount of NH₃ mass produced if 11 grams of N₂ are used, it is possible to use a rule of three knowing the stoichiometry of the reaction. The rule of three used is the following: if 28 grams of N₂ produce 34 grams of NH₃, 11 grams of N₂ how many grams of NH₃ will they produce?

[tex]gramsofNH_{3} =\frac{11gramsofN_{2} *34gramsofNH_{3} }{28 grams of N_{2} }[/tex]

grams of NH₃=13.35

Then, 13.35 grams of NH₃ is made if 11 grams of N₂ are used.

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