CaF2 + (NH4)2O --> 2 NH4F + CaO How many grams of NH4F can be formed from 34.6 grams of CaF2? (AKS 4f)

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 moleCaO: 1 mole

Being the molar mass of the compounds:

CaF₂: 78 g/mole(NH₄)₂O: 52 g/moleNH₄F: 37 g/moleCaO: 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 gNH₄F: 2 mole* 37 g/mole= 74 gCaO: 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

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

Naoh + hbr -> nabr (salt) + h2o ca (oh)2 + 2hbr - - - > ca (br)2 + 2h2o

It must be measured in k.hope my answer is to you : )