BaF2 would be much the best electrical conductor in the list, because it is the only one that is ionically bonded. When molten, such compounds are excellent conductors.
Assuming that you are combining the P4 with Cl2 and there is a suffiecient quantity of Cl2 for the P4 to completely react, you will first need a balanced equation which is P4 + 10Cl2 -> 4PCl5. From there, it's mostly stoichiometry. Take the 24g of P4, divide by the molar mass (123.88g/mol) to get the number of moles of P4 that you have (0.194). You then have to convert, using the balanced equation, from moles of P4 to moles of PCl5, in this case multiplying by 4. That will give you the number of moles of PCl5. The stoichiometry should look something like this 24.0 g P4 x (1 mol P4/123.88g P4) x (4 mol PCl5/1 mol P4).
What mass of will be produced from the given masses of both reactants? 28.0 g/P4 / 123.9 g/P4 = 0.2260 moles/ P4 0.2260 moles /P4 * 4 moles PCl5/1 mole/P4=0.904 moles/PCl5 54.0 g/Cl2 / 70.9 g/Cl2 = .7616 moles/Cl2 .7616 moles / Cl2 *4 moles PCl5/10 moles Cl2=.30 This is about limiting reagents You need to use 2P + 5Cl2 --> 2PCl5 [or P4 + 10Cl2 --> 4PCl5 if you prefer] that tells you that 2P = 62g needs 5Cl2 = 355g Cl2 to react 69.3g