Hydrogen fluoride is used in the manufacture of freons (which destroy ozone in the stratosphere) and in the production of aluminum metal. it is prepared by the reaction caf2 + h2so4 → caso4 + 2hf in one process, 6.25 kg of caf2 is treated with an excess of h2so4 and yields 2.35 kg of hf. calculate the percent yield of hf. % yield

The percentage yield of HF is 73.36 %

Explanation:

To calculate the number of moles, we use the equation:

    ....(1)  

For calcium fluoride:

Given mass of calcium fluoride = 6.25 kg  = 6250 g   (Conversion factor: 1 kg = 1000 g)

Molar mass of calcium fluoride = 78.07 g/mol

Putting values in above equation, we get:  

For the given chemical reaction:

By Stoichiometry of the reaction:

1 mole of calcium fluoride produces 2 moles of hydrofluoric acid

So, 80.05 moles of calcium fluoride will produce = of hydrofluoric acid

Now, calculating the theoretical yield of hydrofluoric acid using equation 1, we get:

Moles of of hydrofluoric acid = 160.1 moles

Molar mass of hydrofluoric acid = 20.01 g/mol

Putting values in equation 1, we get:

To calculate the percentage yield of hydrofluoric acid, we use the equation:

Experimental yield of hydrofluoric acid = 2.35 kg

Theoretical yield of hydrofluoric acid = 3.20 kg

Putting values in above equation, we get:

Hence, the percentage yield of HF is 73.36 %

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.  

the answer is minerals.