Chemistry, 15.10.2019 18:00 nev322

Freon−12 (cf2cl2), widely used as a refrigerant and aerosol propellant, is a dangerous air pollutant. in the troposphere, it traps heat 25 times as effectively as co2, and in the stratosphere, it participates in the breakdown of ozone. freon−12 is prepared industrially by reaction of gaseous carbon tetrachloride with hydrogen fluoride. hydrogen chloride gas also forms. how many grams of carbon tetrachloride are required for the production of 28.5 dm3 of freon−12 at 21°c and 1.62 atm?

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Answer from: lemmeboiz43

The mass of carbon tetrachloride required is 293.8 grams.

Explanation:

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

PV=nRT

where,

P = pressure of Freon-12 = 1.62 atm

V = Volume of Freon-12 = 28.5dm^3=28.5L (Conversion factor: 1dm^3=1L )

T = Temperature of Freon-12 = 21^oC=[21+273]K=294K

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles Freon-12 = ?

Putting values in above equation, we get:

$1.62atm\times 28.5L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 294K\\n=\frac{1.62\times 28.5}{0.0821\times 294}=1.91mol$

The chemical equation for the formation of Freon-12 follows:

$CCl_4+2HF\rightarrow CF_2Cl_2+2HCl$

By Stoichiometry of the reaction:

1 mole of Freon-12 is formed by 1 mole of carbon tetrachloride

So, 1.91 moles of Freon-12 will be formed by = $\frac{1}{1}\times 1.91=1.91mol$ of carbon tetrachloride

To calculate the mass of carbon tetrachloride, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of carbon tetrachloride = 153.82 g/mol

Moles of carbon tetrachloride = 1.91 moles

Putting values in above equation, we get:

$1.91mol=\frac{\text{Mass of carbon tetrachloride}}{153.82g/mol}\\\\\text{Mass of carbon tetrachloride}=(1.91mol\times 153.82g/mol)=293.8g$

Hence, the mass of carbon tetrachloride required is 293.8 grams.

Answer from: Quest

answer : (a) volume of the gas = 8.67 × $10^{-3}$ l

(b) second virial coefficient b = -1.21226 l/mole

solution : given, t = 300k

p = 20atm

z = 0.86

n = 8.2 mmoles = 8.2 × $10^{-3}$ moles

r = 0.082 l atm/k/mol

find, (a) volume of gas and (b) second virial coefficient b at 300k.

formula used : $z=\frac{pv}{nrt}$

and $z=1+\frac{b}{v}$

(a) $z=\frac{pv}{nrt}$

$v=\frac{znrt}{p}$

= $\frac{0.86\times 8.2\times[tex]10^{-3}$ \times 0.082\times 300}{20}[/tex]

= 8.67 × $10^{-3}$ l

(b) $z=1+\frac{b}{v}$

$b=(z-1)v$

= (0.00867-1) × 8.659

= - 8.5839 l/mole

Answer from: Quest

Combustion is the act or process of burning. therefore, i believe the answer is a: burning wood.

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Freon−12 (cf2cl2), widely used as a refrigerant and aerosol propellant, is a dangerous air pollutant...

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