6 moles of` N_2(g)`and 9 moles of `H_2(g)`are allowed to react in a closed rigid vessel at TK temperature . Calculate volume of` NH_3(g)` formed at STP, if % yield 0f reaction is 50% .

9 moles of "D" and 14 moles of E are allowed to react in a closed vessel according to given reactions. Calculate number of moles of B formed in the end of reaction, if 4 moles of G are present in reaction vessel. (percentage yield of reaction is mentioned in the reaction) Step -1 3D+4E 80%rarr5C+A Step-2 3C+5G 50%rarr 6B+F

n mole of PCl_(3) and n mole of Cl_(2) are allowed to react at constant temperature T to have a total equilibrium pressure P , as : PCl_(3)(g)+Cl_(2)(g) hArr PCl_(5)(g) If y mole of PCl_(5) are formed at equilibrium , find K_(p) for the given reaction .

A mixture of 2 moles of N_2,2.0 moles of H_2 and 10.0 moles of NH_3 is introduced into a 20.0 L reaction vessel at 500 K . At this temperature, equilibrium constant K_c is 1.7 xx 10^2 , for the reaction N_2(g) +3H_2(g) hArr 2NH_3(g) (i) is the reaction mixture at equilibrium ? (ii) if not, what is the direction of the reaction ?

4 moles each of SO_(2) "and" O_(2) gases are allowed to react to form SO_(3) in a closed vessel. At equilibrium 25% of O_(2) is used up. The total number of moles of all the gases at equilibrium is

2 mol of N_(2) is mixed with 6 mol of H_(2) in a closed vessel of 1L capacity. If 50% of N_(2) is converted into NH_(3) at equilibrium, the value of K_(C ) for the reaction N_(2)(g0+3H_(2)(g)hArr 2NH_(3)(g) is