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Chemistry HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!?
Imagine that you have a 6.50 L gas tank and a 2.50 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 145 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.
2 Answers
- Dr WLv 71 decade agoFavorite Answer
2 CH≡CH + 5 O2 ---> 4 CO2 + 2 H2O... right?
PV=nRT
nO2 = PO2 x VO2 /(R x TO2)
from the balanced equation....
nO2 x (2 moles ethylene / 5 moles O2) = moles ethylene.. right?
multiplying the above equation for moles O2 by 2/5 we get...
nC2H4 = 2/5 x PO2 x VO2 /(R x TO2)
and
PV = nRT
rearranging..
P = nRT/V
so that for ethylene...
PC2H4 = nC2H4 x R x TC2H4 / VC2H4
substituting...
PC2H4 = [2/5 x PO2 x VO2 /(R x TO2)] x R x TC2H4 / VC2H4
if T is the same in both tanks.. then T cancels out.. right?.. if T isn't constant, you can't solve this.. so let's assume it is...then...
PC2H4 = 2/5 x PO2 x VO2 / VC2H4 = 2/5 x 145 atm x 6.50 L / 2.50 L = 151 atm
- 1 decade ago
i need to know the reaction, assuming that with every 5 mol usage of oxygen we use 2 mol acetylene, then the solution is this:
PV=nRT
for oxygen: 145*6.5=8.314*(nT) --> nT=113.36
the nT term for acetylene should be equal(we assumed that the temprture is equal in both tanks), in fact PV for both gases should be equal coz R is constant(if the stiokumetry is the same) then the moles would be equal, but by using 5 mol oxygen we use 2 mol acetylen, so the proportion is 2/5, therefor:
P=(2/5)*(6.5*145)/2.5=150.8 atm
