If `1:3` (molar ratio) mixture of `N_2` and `H_2` yields 20% (by volume) of `NH_3` at `30 atm`, then the moles of `N_2` converted into the product at equilibrium will be

If 340 g of a mixture of N_2 and H_2 in the correct ratio has a 20% yield of NH_3 . The mass produced would be

If 340 g of a mixture of N_(2) and H_(2) in the correct ratio gas a 20% yield of NH_(3) . The mass produced would be:

When a mixture of N_2 and H_2 in the volume ratio of 1:5 is allowed to react at 100 k and 10^3 atm pressure , 0.426 mole fraction of NH_3 is formed at equilibrium . The K_p for the reaction is

When a mixture of N_2 and H_2 in the volume ratio of 1:5 is allowed to react at 100 k and 10^3 atm pressure , 0.426 mole fraction of NH_3 is formed at equilibrium . The K_p for the reaction is

If 340 g of mixture N_2 and H_2 in the correct raito gave a 20% yield of NH_3 . The mass produced would be:

A mixture containing N_2 and H_2 in a mole ratio 1 : 3 is allowed to attain equilibrium when 50% of the mixture has reacted. If P is the pressure at equilibrium, then the partial pressure of NH_3 formed is