N_2(g)+3H_2(g)->NH_3(l)
Stochiometric ratio of the above raction suggests that 1mole (28g) N_2(g) reacts with 3moles (3xx2=6g)" "H_2(g) to produce 2moles (2xx17=34g)" "NH_3(l)
50g of each reactant is taken . So it is obvious that N_2 will be limilting reactant , hence will be consumed completely to give the product.
By the balanced equation,28g N_2(g) reacts with 6g H_2(g) to produce 34g NH_3(l)
So 50 g N_2(g) reacts completely with 6/28xx50g~~
10.71g" "H_2(g) leaving (50-10.71)=39.29g" "H_2(g)
Now we are to calculate the volume (V) of the remaining w=39.29g" "H_2(g) at pressure
P=760mm=1atm and temperature T=200K
Now by equation of ideal gas we have
PV=w/MRT
here M=2g/"mol"
R=0.082LatmJ^-1mol^-1
So
V=w/Mxx(RT)/P
=>V=39.29/2xx(0.082xx200)/1L=322.178L
