Thermos chemistry-enthalpy question

Question

The value of delta H• for the reaction below is -790kJ. The enthalpy change accompanying the reaction of 0.95g of A is ------kJ. 2S(s)+3O2(g)--> 2SO3

Answer is 23

but I chose 11.7 because

(Here is what I did)

0.95gx 1mol/32gx790kJ/2mol=11.7kJ

Why would you not divide 2 when there is 2moles of S(s)?

Answer 1

The error in your solution is that the unit for ΔH° is not expressed correctly. ΔH° is expressed in units of J/mol or kJ/mol.

From the given mass of sulfur (0.95 g), you divide by its molar mass to obtain moles of sulfur as you did in your solution.

Then the ΔH (enthalpy change) for the above reaction = 0.95 g x (1 mol / 32 g) x -790 kJ/mol = -23 kJ

As a result, you do not have to divide by 2 moles using the coefficient from the balanced chemical reaction.