Calculate the enthalpy change in kJ when 6.32 g P reacts with 14.2 liquid Br2

Question

Red phosphorus reacts with liquid bromina according to the thermochemical reaction. Calculate the enthalpy change in kJ when 6.32 g P reacts with 14.2 liquid Br₂

2P (s) + 3Br₂(l) --> 2PBr₃ (g) ΔH_rxn = -246 kJ

Answer 1

2P (s) + 3Br₂(l) ==> 2PBr₃ (g) ΔH_rxn = -246 kJ

Since we are given the amounts of both reactants, we must first determine which, if either, reactant is in limting supply. One easy way to do this is to simply divide the moles of each by the corresponding coefficient in the balanced equation, and which ever value is less represents the limiting reactant:

For P: 6.32 g P x 1 mol P / 30.97 g = 0.204 moles P (÷2->~0.1)

For Br₂: 14.2 g Br₂ x 1 mol Br₂ / 159.8 g = 0.0889 moles Br₂ (÷3->~0.03)

Since 0.03 is less than 0.1, Br₂ is the limiting reactant, and we will use 0.0889 moles Br₂.

∆H = 0.0889 mols Br₂ x -246 kJ / 3 mols Br₂

∆H = -7.29 kJ