Hydrogen gas, H2, reacts explosively with gaseous chlorine, Cl2, to form hydrogen chloride, HCl(g). What is the enthalpy change for the reaction of 2 mole of H2(g) with 2 mole of Cl2(g) if both the reactants and products are at standard state conditions? The standard enthalpy of formation of HCl(g) is −92.3 kJ/mol. H2(g)+Cl2(g)→2HCl(g)

Given :

Hydrogen gas, , reacts explosively with gaseous chlorine, , to form hydrogen chloride, HCl(g).

The standard enthalpy of formation of HCl(g) is −92.3 kJ/mol.

To Find :

The enthalpy change for the reaction of 2 mole of with 2 mole of if both the reactants and products are at standard state conditions .

Solution :

We know ,

( Since , and are their most elementary form . So their enthalpy of formation is 0 kJ )

Therefore , the enthalpy change for the reaction of 2 mole of with 2 mole of if both the reactants and products are at standard state conditions is -184.6 kJ/mol .

Hence , this is the required solution .

calculate how many grams of ammonia can be produced from 4.21 grams of

n2 + 3h2 = 2nh3

my answer:

25.0g n2 x (1 mol/ 28.0gn2) x (2 mol nh3/ 1 mol n2) x (17.0g nh3/ 1 molnh3) = 30.36g n2

25.0g h2 x (1 mol h2/ 2.00g h2) x (2 mol nh3/ 3 mol h2) x (17.0g nh3/ 1 mol nh3) = 141.67g h2

lr = n2

narendra modi is currently the pm of india

explanation: