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?
mass of NH₄F= 32.83 grams
<u><em>32.83 grams of NH₄F can be formed from 34.6 grams of CaF₂</em></u>
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