Imagine that you have a 7.00L gas tank and a 3.50L gas tank. You need to fill one tank with oxygen and the other with acetylene to use in conjunction with your welding torch. If you fill the larger tank with oxygen to a pressure of 135atm , to what pressure should you fill the acetylene tank to ensure that you run out of each gas at the same time? Assume ideal behavior for all gases.

I don't know the answer, I don't know how to do it, and your answer didn't get me any closer to solving it.

I didn't work it for you but I gave you the three steps to solve it. It isn't easy to help if you don't tell me what you don't understand.

Three steps. Copied from above.

1. PV = nRT for the 7 L tank for O2 @ 135 atm and solve for n.
2. Use the equation to convert mols O2 to mols C2H2.
3. Use PV = nRT and n mols C2H2 and solve for P in the smaller tank.

I'll start.
Step 1 from above.
PV = nRT
P = 135 atm
V = 7 L
n = ?
R = 0.08206
T = isn't given so use any T but use the same T throughout. An easy choice would be 300 K.
135*7 = n*0.08206*300.
Solve for n. You do that first.
What is n?

Good.

Now use the coefficients in the balanced equation to convert mols O2 to mols C2H2. That is
38.38 mols O2 x (2 mols C2H2/5 mols O2) = 38.38 mols x 2/5 = about 15 but you can clean up the numbers.
Now use the smaller tank and PV = nRT and solve for P.
P = ?
V = 3.50 L
n = about 15 but use your more accurate number.
R = same as above.
T = 300 if that's what you used before.

2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g) formula for above

To find the pressure at which the acetylene tank should be filled, we need to consider the ideal gas law, which states:

PV = nRT

Where:
P = pressure
V = volume
n = number of moles of gas
R = ideal gas constant
T = temperature (assumed to be constant)

We can rearrange the equation to solve for the number of moles of gas:

n = PV / RT

Since we want to run out of oxygen and acetylene at the same time, we can equate the number of moles of each gas:

n(oxygen) = n(acetylene)

Let's assume the temperature and the amount of gas in moles are the same for both tanks. We can equate the two expressions for the number of moles:

P(oxygen) * V(oxygen) / RT = P(acetylene) * V(acetylene) / RT

Since the temperature and the gas constant are constant, we can simplify the equation:

P(oxygen) * V(oxygen) = P(acetylene) * V(acetylene)

Now, we can substitute the given values:

P(oxygen) = 135 atm
V(oxygen) = 7.00 L
V(acetylene) = 3.50 L

Plugging these values into the equation, we can solve for P(acetylene):

135 atm * 7.00 L = P(acetylene) * 3.50 L

945 atm * L = P(acetylene) * 3.50 L

P(acetylene) = 945 atm / 3.50

P(acetylene) ≈ 270 atm

Therefore, you should fill the acetylene tank with a pressure of approximately 270 atm to ensure that you run out of both oxygen and acetylene at the same time.

107.95 atm

I think you're allowed only 3 significant figures so I would round that to 108,

Look through your problem and make sure no temperature was given.
I must leave the house. Back in about an hour.

38.37

PV = nRT for the 7 L tank for O2 @ 135 atm and solve for n.

Use the equation to convert mols O2 to mols C2H2.
Use PV = nRT and n mols C2H2 and solve for P in the smaller tank.