( 1)2mol？ N2 and 6mol? H2 is completely converted into 4mol ammonia, and the heat of 184.8kJ is released. Because it is a reversible reaction, the heat released is q <184.8 kj;

A.3v inverse (N2)=v positive (H2), and the ratio of reaction rates is equal to the stoichiometric ratio, so v inverse =v positive, so a is correct;

B.2v positive (H2)=v positive (NH3), and the reaction rate is different from the stoichiometric ratio, so b is wrong;

C. Density = total mass volume, the total mass is constant, and the volume is constant, so the density of the mixed gas is constant, which cannot be used as a basis for judging whether the chemical equilibrium state is reached, so C is wrong;

D. The concentration ratio of each substance depends on the amount and conversion rate of the starting substance of the substance, so c (N2): c (H2): c (NH3) =1:3: 2 can't be used as the basis for judging whether the equilibrium state is reached, so d is wrong.

So the answer is:

(2) With the increase of C (NH3), the equilibrium moves forward, so the conversion rate of CO2 increases.

So the answer is: c(NH3) increases and the equilibrium moves forward, so the conversion rate of CO2 increases;

(3)2NH3~CO2

Assuming that CO2 is 2mol at the beginning, ammonia is 2mol× 4 = 8amol, CO2 is 2mol× 0.5 = amol at equilibrium, converted CO2 is 2mol-amol = amol, and converted ammonia is amol×2=2amol, the conversion rate of ammonia is 2mol 8amol× 100% = 25%.

So the answer is: 25%;

(4)3Cl2+2NH3→N2+6HCl①

3Cl2+8NH3→N2+6NH4Cl②

12Cl2+ 15NH3→, Cl decreased from 0 to-1, * * decreased by 24, n increased from -3 to 0, and * * increased by 24, so 8 ammonia gases were oxidized to generate nitrogen, and 7 ammonia gases were not oxidized, so the product was 4N2+7NH4 Cl.

So the answer is:12cl2+15n3 = 4n2+7NH4Cl+17hcl.