Alright, so we are going to use the ideal gas law to solve this one. That law simply states that

#PV=nRT#

In this equation, #P# is pressure in atm

#V# is volume in liters

#n# is the amount of gas in moles

#R# is a constant equal to #(0.08206L*atm)/(K*mol)#

And finally, #T# is temperature in kelvin

So we are trying to solve for #P#, so we need that on one side of the equation. We can do that by just dividing by #V#

#(PcancelV)/cancelV=(nRT)/V#

#P=(nRT)/V#

Ok, at this point lets start plugging variables in.

#V=10.0L#

#P=(nRT)/(10.0L)#

To convert #C^o to K# we just add 273

#K=C^o +273#

#K=200^oC+273#

#T=473K#

#P=(n*R*473K)/(10.0L)#

Now for the tricky part. We need to convert 32g of #C_2H_6# into moles. To do this we must first find the molar mass of #C_2H_6#, then convert 32g to moles.

The molar mass is going to equal #(30.07g)/(mol)# because we simply add the atomic mass of both elements:

#C=12.0107*2#

#H=1.00794 *6#

....................................

#(30.07g)/(mol)#

So now we need to find out how much moles 32g is

#n=(mol)/(30.07g)*32g#

#n=(mol)/(30.07cancelg)*32cancelg#

#n=(32mol)/(30.07)#

#n=1.1mol#

Ok, lets plug that in and the constant #(R)#. I'd like to note that I like to put the constant outside the equations as it makes it a lot easier.

#P=(n*R*473K)/(10.0L)#

#P=(1.1mol*473K)/(10.0L)(0.08206L*atm)/(K*mol)#

This is the fun part, were you cancel out all the conversion names. NOTE: since you are looking to #P#, we should have #atm# left over at the end

#P=(1.1cancel(mol)*473cancelK)/(10.0cancelL)(0.08206cancelL*atm)/(cancelK*cancel(mol))#

Now, finally, we just solve! Notice that atm is all that left....

#P=(1.1*473*0.08206*atm)/(10.0)#

#P=4.70atm#

Hope this helped! If you have any more questions feel free to ask them!

~Chandler Dowd