#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#