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?
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.