Question

what is the delta H°reaction for the reaction shown below when 54.2g of Cl2 react? P4(s) + 6Cl2(g) ---> 4PCl3(g) delta H = -1226kj

Answer 1

H gives us the change in enthalpy of a reaction.  H is dependent on the moles of product formed in the reaction.

The balanced chemical reaction is given to us as -

P4 + 6Cl2 -> 4PCl3 and its H = -1226 kJ

As we can see this H value corresponds to the formation of 4 moles of PCl3 from 6 moles of Cl2

so, for every 6 moles of Cl2 4 moles of PCl3 we will be formed so, 1 mole of Cl2 will form 4/6 moles of PCl3 and thus  H = 4*(-1226kJ)/6

But we have 54.2 g of Cl2 which is 54.2 g/ 71g/mol = 0.7633 moles as moles = weight/molar mass

so, 0.7633 moles will produce 0.7633*4/6 moles of PCl3 and  H = 0.7633*4*(-1226kJ)/6 = -623.936 kJ