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Hydrogen chloride gas is shipped in a container under 5,100 mmHg of pressure that occupies 20.1 liters at 29°C. How many liters of gas would be produced at STP?

Respuesta :

Hagrid
This problem is solved using the ideal gas equation. To simplify the problem, we must assume that the gas is ideal then proceed with the ideal gas equation. PV=nRT. First, get the number of moles of the gas. With R = 0.0821 L atm/ mol K and 5100mmHg = 6.7atm. Plug in the given resulting to an answer n=5.4 moles. 

At STP, regardless of the gas specie the molar volume is always 22.4L/mol. The final answer is then 22.4*5.4 = 121.L

Answer:

121.92 liters of gas would be produced at STP.

Explanation:

The combined gas equation is,

[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]

where,

[tex]P_1[/tex] = initial pressure of Hydrogen chloride gas = 5,100 mmHg

(1 atm = 760mmHg)

[tex]=\frac{5,100}{760} atm =6.71 atm[/tex]

[tex]P_2[/tex] = final pressure of Hydrogen chloride gas =  1atm (STP)

[tex]V_1[/tex] = initial volume of Hydrogen chloride gas = [tex]20.1 L[/tex]

[tex]V_2[/tex] = final volume of Hydrogen chloride gas = ?

[tex]T_1[/tex] = initial temperature of Hydrogen chloride gas = [tex]29^oC=273+29=302K[/tex]

[tex]T_2[/tex] = final temperature of Hydrogen chloride gas = [tex]0^oC=273+0=273 K[/tex]  

Now put all the given values in the above equation, we get:

[tex]\frac{6.71 atm\times 20.1 L}{302K}=\frac{1 atm\times V_2}{273K}[/tex]

[tex]V_2=121.92 L[/tex]

121.92 liters of gas would be produced at STP.

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