A sample containing 6.40 g O2 gas has a volume of 15.0 L. Pressure and temperature remain constant.

a. What is the new volume if 0.600 mole O2 gas is added?
b. Oxygen is released until the volume is 12.0L. How many moles of O2 are removed?
c. What is the volume after 7.50 g He is added to the O2 gas already in the container? plss neeed help

a. I would use PV = nRT and solve for P (that gives you initial P). No T is listed but you can use any convenient T. Add mols O2 initially + 0.6 mol O2 and use PV = nRT again to solve for new volume. Use the old P you calculated and don't change T from what you used.

b. Redo PV = nRT with new V = 12.0 L. Solve for n O2 and subtract from total mols O2 to find moles removed.

c. Convert 7.5g He t mols, add to mols O2 available from part b and redo PV = nRT and solve for V.

in a. I got 60 L for volume and .8 moles(.2+.6moles. now in b. i got .16 mole so where do I subtract .16 from should I subtract it from .8 moles or from the initial mole which is .2 mole

From the 0.8; that's where the volume was 60 L.

By the way there is a shorter way of doing this
Notice that if we use PV = nRT and solve for n = (PV/RT). Since P, R, and T are constant, the n = V; therefore,
(n1/V1) = (n2/V2). As an eample, for part a, (0.2/15) = (0.8/V2) and V2 = 60 which can be done in your head.
For part b, (0.8/60) = (n2/12); n2 = 0.16

To solve these problems, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

a. To find the new volume when 0.600 mole O2 gas is added, we first need to know the initial number of moles. We are given that the initial sample contains 6.40 g of O2 gas. To convert grams to moles, we can use the molar mass of O2, which is 32.00 g/mol.

Initial moles of O2 = 6.40 g O2 / 32.00 g/mol = 0.200 mole O2

Now, we can find the total number of moles after adding 0.600 mole O2:

Total moles of O2 = 0.200 mole + 0.600 mole = 0.800 mole

Since the pressure and temperature remain constant, we can use the ideal gas law to find the new volume:

(Initial pressure) * (Initial volume) = (Final pressure) * (Final volume)

Rearranging the equation to solve for the final volume:

Final volume = (Initial pressure) * (Initial volume) / (Final pressure)

Since the pressure is constant, we can cancel it out:

Final volume = (Initial volume) * (Total moles of O2) / (Initial moles of O2)

Plugging in the values we found:

Final volume = 15.0 L * 0.800 mole / 0.200 mole = 60.0 L

So, the new volume after adding 0.600 mole O2 gas is 60.0 L.

b. To find the number of moles of O2 removed when the volume is decreased to 12.0 L, we can use the same approach as in part a.

Using the ideal gas law, we can write:

(Initial pressure) * (Initial volume) = (Final pressure) * (Final volume)

Rearranging the equation to solve for the final moles:

Final moles of O2 = (Initial pressure) * (Initial volume) / (Final pressure) * (Final volume)

Since the pressure remains constant, we can cancel it out:

Final moles of O2 = (Initial volume) / (Final volume) * (Initial moles of O2)

Plugging in the known values:

Final moles of O2 = 15.0 L / 12.0 L * 0.200 mole = 0.250 mole

Therefore, 0.250 mole of O2 gas is removed when the volume is decreased to 12.0 L.

c. To find the final volume after adding 7.50 g of He to the O2 gas, we need to consider both gases separately.

First, we calculate the number of moles of O2 using the given information of 6.40 g:

Moles of O2 = 6.40 g / 32.00 g/mol = 0.200 mole

Next, we calculate the number of moles of He using its molar mass, which is 4.00 g/mol:

Moles of He = 7.50 g / 4.00 g/mol = 1.875 mole

Now, we can add the moles of O2 and He together to get the total moles:

Total moles = Moles of O2 + Moles of He = 0.200 mole + 1.875 mole = 2.075 mole

Since the pressure and temperature remain constant, we can use the ideal gas law to calculate the final volume:

Final volume = (Initial pressure) * (Initial volume) / (Total moles)

Plugging in the values we found:

Final volume = 15.0 L * 0.200 mole / 2.075 mole = 1.452 L

So, the final volume after adding 7.50 g of He to the O2 gas is 1.452 L.