Question

What is the mass and # of molecules of PCl3 formed in the reaction of 75.0 g P4 with 275 g Cl2 ?

Answer 1

The mass of PCl₃ is 332.3 g.

The molecules of PCl₃ is
`1.45*10^(24)\ g`

Step-by-step explanation:

Given that,

Mass of P₄ = 75.0 g

Mass of Cl₂ = 275 g

We know that,

The reaction is

`P_(4)+6Cl_(2)\Rightarrow 4PCl_(3)`

We need to calculate the moles of P₄

Using formula of moles

`n=(m)/(M)`

Put the value into the formula

`n=(75.0)/(123.89)`

`n=0.61\ moles`

We need to calculate the moles of Cl₂

Using formula of moles

`n=(m)/(M)`

Put the value into the formula

`n=(275)/(70.906)`

`n=3.89 moles`

We need to calculate the number of moles of
`4PCl_(3)`

Moles in 1 P₄ = 0.61

So,
`moles\ in\ 4PCl_(3) = 4*0.61`

`moles\ in\ 4PCl_(3)=2.42\ moles`

We need to calculate the mass of
`PCl_(3)`

Using formula of mass

`m=M* n`

Put the value into the formula

`m=137.3*2.42`

`m=332.3\ g`

We need to calculate the molecules of
`PCl_(3)`

Using formula of molecules

`molecules=n* N`

Where, n = number of moles

N = avagadro number

Put the value into the formula

`molecules=2.42*6.023*10^(23)`

`molecules=1.45*10^(24)\ g`

Hence, The mass of PCl₃ is 332.3 g.

The molecules of PCl₃ is
`1.45*10^(24)\ g`