A mixture consists of 28% oxygen, 14% hydrogen, and 58% nitrogen by volume. A sample of this mixture has a pressure of 4.0 atm in a 9.6 L container at 300 K temperature. This mixture is run through a gas filter, which removes the hydrogen from this system. The gas is placed back into the 9.6 L container. How many moles of gas are left in the container? A) 4.2 mol B) 2.6 mol C) 1.3 mol D) 3.9 mol

C) 1.3 mol

Explanation:

Using gas law we can find the initial moles of the sample of the mixture, as follows:

PV = nRT

PV / RT = n

Where P is pressure: 4.0atm

V is volume: 9.6L

R is gas constant: 0.082atmL/molK

T is absolute temperature: 300K

And n are moles of the gas

PV / RT = n

4.0atm*9.6L / 0.082atmL/molK300K = n

n = 1.56moles of the mixture of the gas are present into the 9.6L container

Now, 14% of this gas is hydrogen that was removed of the system, that is:

1.56mol*14% = 0.22 moles of hydrogen are removed.

Thus, moles of gas that remains in the container are:

1.56mol - 0.22mol = 1.34mol.

Right answer is: