Respuesta :

CintiaSalazar

Answer:

32.83 grams of NH₄F can be formed from 34.6 grams of CaF₂

Explanation:

The balanced reaction is:

CaF₂ + (NH₄)₂O ⇒ 2 NH₄F + CaO

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following quantities participate in the reaction:

  • CaF₂: 1 mole
  • (NH₄)₂O: 1 mole
  • NH₄F: 2 mole
  • CaO: 1 mole

Being the molar mass of the compounds:

  • CaF₂: 78 g/mole
  • (NH₄)₂O: 52 g/mole
  • NH₄F: 37 g/mole
  • CaO: 56 g/mole

Then by stoichiometry, the following amounts of mass participate in the reaction:

  • CaF₂: 1 mole* 78 g/mole= 78 g
  • (NH₄)₂O: 1 mole* 52 g/mole= 52 g
  • NH₄F: 2 mole* 37 g/mole= 74 g
  • CaO: 1 mole* 56 g/mole= 56 g

You can apply the following rule of three: if 78 grams of CaF₂ form 74 grams of NH₄F by stoichiometry, 34.6 grams of CaF₂ will form how much mass of NH₄F?

[tex]mass of NH_{4} F=\frac{34.6 grams of CaF_{2} *74 grams ofNH_{4} F}{78grams of CaF_{2} }[/tex]

mass of NH₄F= 32.83 grams

32.83 grams of NH₄F can be formed from 34.6 grams of CaF₂

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