Question :
4 moles of A(g) is added to a fixed volume container and left until they arrive at the equilibrium status.
2A(g) <===> A2(g)
At 127 C total pressure of the equilibrium system is 8.314*10^5 Pa and the density of the mixture of gas is 10 kg/m^3 . If the relative atomic weight of A is 30 find the partial pressure of A2(g).
My first attempt towards solving this question is applying PM=dRT equation, where I took d as (30+x) (x is the relative atomic mass of A2(g) )
Then I got x=220
My next thought is finding moles of A2(g) and then obtaining the partial pressure of it.Then again I'm getting another unknown value for moles of A2(g) as follows;
2A(g) <===> A2(g)
initial(moles) 4 -
final (moles) (4-2y) y
How do I solve this?
2 answers