Question

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?

Answer 1

121.92 liters of gas would be produced at STP.

Step-by-step explanation:

The combined gas equation is,

`(P_1V_1)/(T_1)=(P_2V_2)/(T_2)`

where,

`P_1` = initial pressure of Hydrogen chloride gas = 5,100 mmHg

(1 atm = 760mmHg)

`=(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:

`(6.71 atm* 20.1 L)/(302K)=(1 atm* V_2)/(273K)`

`V_2=121.92 L`

121.92 liters of gas would be produced at STP.

Answer 2

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