A rigid, closed container contains 0.0700 mol of Ne(g) and a sample of the solid

Question

A rigid, closed container contains 0.0700 mol of Ne(g) and a sample of the solid, ammonium nitirite, NH4NO2. Assume that the volumes of solids are negligible compared to the volume of the container. The pressure of the Ne is meausre at 33 degrees celsius and is found to be .409atm.

The container is then heated to 333 degrees celsius and all of the ammonium nitrite decomposes according to the reaction. NH4NO2(s) --> N2(g) + 2H2O(g)

The Ne is not involved in the reastion and is still present. The final pressure in the container after complete decompostion of the NH4NO2 is 3.93 atm. Assuming ideal gas behavior, answer the following questions.

a. What is the volume of the container, in liters? ____L

b. What is the partial pressure of H2O in the containers, at 333 C, when the reaction is complete?

pH2O=____atm

Explanation / Answer

if rigid, volume won't change, so

mol of Ne = 0.07

T = 33°C = 306 K

P = 0.409 atm of Ne

then, after heating

Tnew = 333°C = 606 K

final P = 3.93 atm , includes Ne....

a)

intially, we only have Neon, so we can assume that the volume occupied by it is the 100% of container

PV = nRT

V = nRT/P

n = 0.07, R =0.082 Latm/molK P = 0.409 atm, T = 306 K

so

V = (0.07)(0.082)(306)/(0.409 ) = 4.2944 Liters

b)

find partial pressure of H2O at T = 333°C...

so.. there will be 3 gases:

Ne, H2O and Ne...

first, identify the total moles that will occupy that volume at such T/P

PV = nRT

n total = PV/(RT) = (3.93)(4.2944 )/(0.082*606) = 0.33963 mol

note that:

Ne does not react so total mol = 0.33963 - 0.07 mol of ne = 0.26963 mol left

of that.... we know that it is N2 and H2O for sure...

ratio is 1:2, due to stoichiometry

so

0.26963 --> 3x

x = 0.26963/43 = 0.08987 mol

so

N2 = x = 0.08987 mol of N2

H2O = 2x = 2*0.08987 = 0.17974 mol of H2O

then....

P-H2O = x-H2O * P-total

x-H2O = mol of H2O / total mol = 0.17974 / 0.33963 = 0.52922

so..

P-H2O = 0.52922*3.93 atm = 2.079834 atm of H2O

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