Imagine that you have a 6.00 L gas tank and a 3.00 L 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 115 atm, 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.

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1. Write the equation.
2C2H2 + 5O2 ==> 4CO2 + 2H2O

2.Use PV = nRT to determine n (# mols) O2 in the larger tank. No T is given; therefore, pick any T and use that T throughout the calculations.

3. Using the coefficients in the balanced equation, convert mols O2 to mols C2H2.

4. Now use PV = nRT to calculate P of the smaller tank for C2H2.

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2. The T is have to use is 25C,and im trying to find the number of moles of oxygen first, so

PV= nRT
(115 atm)(6L)= n(0.08206)(25)

bUH IM NOT SURE IF I NEED TO CONVERT THE PRESSURE OR LITRES TO SOMETHING ELSE

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3. P is in atmospheres. V is in liters. T is in Kelvin (Kelvin = 273 + C or 298 for 25 C). R is 0.08205 L*atm/mol*K.

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4. OMg thank you soo much

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5. The balanced chemical reaction is:
2C2H2 + 5O2 --> 4CO2 + 2H2O
The mole ratio of C2H2 to O2 is 2 : 5. Rearranging the Ideal Gas Law, we get:
n = PV / RT or
Let n1 = P1V1/RT (moles of C2H2)
and n2 = P2V2/RT (moles of O2
n1/n2 = P1V1/RT * RT/P2V2
Simplifying,
n1/n2 = P1V1 / P2V2
n1/n2 = (P1/P2)(V1/V2)
Substitute:
n1 = moles of C2H2 (acetylene) = 2
n2 = moles of O2 = 5
P1 = pressure of acetylene (unknown)
P2 = pressure of O2
V1 = volume of C2H2
V2 = volume of O2

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