Question

If the HF formed in reaction (i) cannot be reused, calculate the weight of $F_2^{}$ consumed by 1.0g of \[\left( {CF_2^{}} \right)_n^{}\] produced.
A. 2.0 g
B. 2.52g
C. 1.52g
D. 3.0g

Answer 1

Hint: HF means Hydrogen Fluoride (formula: HF) and hydrogen fluoride is a chemical compound. Hydrofluoric acid is formed by treatment of the mineral fluorite $Ca{F_2}$with the help of highly concentrated acid at a temperature of 265 °C these two substances when react then produce hydrogen fluoride. And also formed in this reaction calcium sulfate.
Complete step-by-step answer:

We can also say that hydrogen fluoride is also a weak acid, meaning it can easily be added to water, only produce hydrogen partially, break down and release hydrogen ion when it is added to water.
When we combine a hydrogen atom with a fluorine atom then Hydrogen fluoride is formed and hydrogen fluoride is a molecule. Hydrogen fluoride is a highly valuable molecule in industry, but Hydrogen fluoride is also a very dangerous chemical for biological systems.
It has a molar mass of 20 grams per mole. In other words we can say that one mole of hydrogen fluoride Is equal to the weight approximately 20 grams.
To solve this question, we use the basic theory related to the chemical reaction. As mentioned in the question HF formed in reaction (i) cannot be reused. And we need to focus on what amount of \[\left( {CF_2^{}} \right)_n^{}\] produced when 1g of $F_2^{}$consumed.
From the chemical equations First and Equation second it can be seen that 1mol $\left( {CF_2^{}} \right)_n^{} \equiv $4n mol $CoF_2^{} \equiv 2n$ mol $F_2^{}$
Hence, 50 n g $\left( {CF_2^{}} \right)_n^{}$=76n g $F_2^{}$
Hence 1.0 g $\left( {CF_2^{}} \right)_n^{} = \dfrac{{76}}{{50}} = 1.52gF_2^{}$
Hence 1.52 g of $F_2^{}$ is consumed by 1.0 g of $\left( {CF_2^{}} \right)_n^{}$ produced.
Therefore, option (B) is correct.

Note- The formation of hydrogen fluoride from single, isolated H and F atoms can be explained using standard enthalpy change of formation in the reaction. Under standard conditions, both hydrogen and fluorine exist in a diatomic gaseous state, but after reaction between these two-results hydrogen fluoride HF.