[Kdub]

1. When trinitrotoluene (TNT) decomposes, solid carbon and a mixture of gases is obtained... 2 mol TNT --> 2 mol C + 12 mol CO + 5 mol H2 + 3 mol N2

In an experiment, 4.68g TNT decomposes completely in a rigid 3.25L container. What is the total pressure in the container if the final temperature is 231*C?

moles TNT = mass / molar mass
                = 4.68g / 227.14g/mol
                = 0.0206 mol

PV = nRT
P = (nRT) / V
  = (0.0206mol x 0.08206 x 504.15) / 3.25
  = 0.262 atm

So now I used mole ratios to find moles and pressure of each molecule and got:
Ptotal = 0.262atm(TNT) + 0.262atm(C) + 1.57atm(CO) + 0.655atm(H2) + 0.393atm(N2) = 3.14atm

Does that seem right?


2. It takes 19.9 minutes for a 10.0mL sample of an unknown gas to effuse through a pinhole. A 10.0mL sample of Helium, He, required 4.27 minutes. What is the molar mass of the unknown gas?

Well this is Graham's Law which states t₁t₂= M₁^1/2M₂^1/2

So I found the rate ... 19.9 minutes / 4.27 minutes = 4.66
Therefore Helium effuses at a rate of 4.66 x faster than that of the unknown gas.
We know the molar mass of Helium is 4.00g/mol, but what ever i do now ends up being the wrong answer...what should the equation look like with all the numbers plugged in?

Thanks a bunch,  
 
 

[Borek]

0.262atm(TNT) + 0.262atm(C)

Since when decomposed substance increases pressure? And since when solid carbon increases pressure?

[Kdub]

carbon (s) was my mistake that gets taken out, but i dont understand how you are supposed to have a lower pressure of products over reactants since theres more molecules on the products side thus more gas and more pressure...

[Borek]

i dont understand how you are supposed to have a lower pressure of products over reactants

Where did you get this idea from? The question doesn't contain such information, it just asks for the final pressure (2.62).

[Donaldson Tan]

regarding q(1), I assume the decomposition of TNT does not consume oxygen and the volume of TNT and carbon are negligible to the volume of the container.

in another words, before decomposition, there is 3.25L of air at ambient (RTP) conditions inside the container. The air does not participate in the reaction but the air is still part of the final mixture after all the TNT had decomposed inside the 3.25L container.

regarding q(2), graham's law is given as:
R1/R2 = (M2/M1)^1/2
where R is the rate of diffusion and M is the molar mass