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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?

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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,

\(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\)

where,

\(P_1\) = initial pressure of Hydrogen chloride gas = 5,100 mmHg

(1 atm = 760mmHg)

\(=\frac{5,100}{760} atm =6.71 atm\)

\(P_2\) = final pressure of Hydrogen chloride gas =  1atm (STP)

\(V_1\) = initial volume of Hydrogen chloride gas = \(20.1 L\)

\(V_2\) = final volume of Hydrogen chloride gas = ?

\(T_1\) = initial temperature of Hydrogen chloride gas = \(29^oC=273+29=302K\)

\(T_2\) = final temperature of Hydrogen chloride gas = \(0^oC=273+0=273 K\)  

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

\(\frac{6.71 atm\times 20.1 L}{302K}=\frac{1 atm\times V_2}{273K}\)

\(V_2=121.92 L\)

121.92 liters of gas would be produced at STP.

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