Question

What mass of PCl5 will be produced from the given masses of both reactants?

Answer 1

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=.305

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

so 28g P needs 355/62 x 28g Cl2 = 160.32g

so you don't have enough chlorine so chlorine is the limiting reagent,.

from eqtn 355g Cl2 gives 479gg PCl5
so 54g Cl2 gives 479/355 x 54 = 72.86g

Answer 2

the moles will be formed on the basis of the limiting reagent
and multiply it with molecular mass

Answer 3

P4 + 10Cl2 = 4PCl5

Thus, 1mole of P4 forms 4 moles of PCl5.

Answer 4

The mass of PCl5 produced from the given masses of both: 63.4g