Question

A 12.39 g sample of phosphorus reacts with 40.75 g of chlorine to form only phosphorus trichloride (PCl3). If it is the only product, what mass (g) of PCl3 is formed? A possibly useful molar mass is P = 30.974 g/mol; PCl3 = 137.324 g/mol; Cl = 35.45 g/mol.

Answer 1

The answer is "52.18 g"

Step-by-step explanation:

Phosphorus mass = 12.39

molar mass of
`P_4 = 4 * 30.974`

`= 123.896 \ (g)/(mol)`

Phosphorus moles =
`(mass)/(molar \ mass)`

`= (12.39)/(123.896) \\\\ = 0.10`

mass = 40.75 g

`Cl_2` molar mass
`= 2 * 35.45 \\\\`

`= 70.9 \ (g)/(mol)`

`Cl_2` moles =
`(mass)/(molar \ mass)`

`= (40.75)/(70.9) \\\\ =0.57`

The balanced equation is:
`P_4 + 6 Cl_2 \longrightarrow 4 PCl_3`

In the step1:

Consider reactor limiting the reaction current is less than necessary quantity as per the balanced equation is considered reactor limiting

The equilibrium equation:

`1 \ mole \ of \ P_4 \to 6 \ moles \ Cl_2\\\\0.1 \ mole \ P_4 = ?`

Moles of
`Cl_2` required = 0.6 moles

But only
`0.57`,
`Cl_2` moles are less than the amount needed Thus,
`Cl_2` is reactant restricted

In the step2:

Find
`PCl_3`Theoretical performance
`PCl_3` the amount formed when the reactant is fully restricted

From equilibrium

`6 \ moles\ Cl_ 2 \to 4 \ moles \ PCl_3 \\\\0.57\ moles \ Cl_2 = ?`

`PCl_3` shaped mass =
`0.57 * (4)/(6)`

`= 0.38 \ moles`

`PCl_3` shaped mass =
`moles * molar\ mass`

Mass shaped by
`PCl_3` =
`0.38 * 137.324`

`= 52.18 \ g`