Question

A 12.39 g sample of phosphorus reacts with 42.54 g of chlorine to form only phosphorus trichloride (PCl3). If it is the only product, what mass of PCl3 is formed?

Answer 1

The chemical reaction involved in this process is

$2 P + 3 Cl_2\rightarrow 2 PCl_3$

$moles(n)=\frac{given mass}{mol.mass}$

$moles\quad of\quad P(n_1)=\frac{12.39 g }{30.9738 g /mol}=0.4mol\\ moles\quad of\quad Cl_2(n_2)=\frac{42.54 g }{70.9064 g Cl2/mol}=0.6mol$

0.400015 mole of P would react completely with $0.400015 \times 1.5 = 0.600023$ mol of Cl₂, but there is not that much Cl₂ present, so Cl₂ is the limiting reactant .

The amount PCl₃ is formed

$0.599945 mol Cl_2\times \frac{2}{3} \times (137.3334 g PCl_3) = 54.93 g PCl_3$

Answer 2

2P + 3Cl2 ---> 2PCl3
molar mass:
P = 31 g/mol
Cl2 = 71 g/mol
moles of P = mass / molar mass = 12.39 / 31 = 0.40 mol
moles of Cl2 = 42.54/71 = 0.60 mol
from the eq above, we know that the ratio of P : Cl2 = 2 : 3.
In fact, the molar ratio of P : Cl2 = 0.40 : 0.60 = 2 : 3
Thus, 12.39 g of P exactly reacts with 42.54 g of Cl2, forming PCl3.
Mass of PCl3 = mass of P + mass of Cl2 = 12.39 + 42.54 = 54.93 g