Question

Phosphorous trichloride can be formed via a combination reaction from its elements. Phosphorous is often represented by its empirical formula, P, in chemical equations, as is carbon. If 12.39 g of phosphorous is combined with 42.54 g of chlorine, what mass of phosphorous trichloride could be formed?

Answer 1

`\text{54.92 g PCl}_(3)}`

Step-by-step explanation:

We are given the masses of two reactants and asked to determine the mass of the product.

This looks like a limiting reactant problem.

1. Assemble the information

We will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.

MM: 30.97 70.91 137.33

2P + 3Cl₂ ⟶ 2PCl₃

Mass/g: 12.39 42.54

2. Calculate the moles of each reactant

`\text{Moles of P} = \text{12.39 g P} * \frac{\text{1 mol P}}{\text{127.33 mol P}} = \text{0.4006 mol P}\\\\\text{Moles of Cl$_(2)$} = \text{42.54 g Cl}_(2) * \frac{\text{1 mol Cl$_(2)$}}{\text{70.91 g Cl$_(2)$}} = \text{0.5999 mol Cl$_(2)$}`

3. Calculate the moles of PCl₃ from each reactant

`\textbf{From P:}\\\text{Moles of PCl$_(3)$} =  \text{0.4006 mol P} * \frac{\text{2 mol PCl$_(3)$}}{\text{2 mol P}} = \text{0.4006 mol PCl$_(3)$}\\\textbf{From Cl$_(2)$:}\\\text{Moles of PCl$_(3)$} =\text{0.5999 mol Cl$_(2)$} * \frac{\text{2 mol PCl$_(3)$}}{\text{3 mol Cl$_(2)$}} = \text{0.3999 mol PCl$_(3)$}`

4. Identify the limiting reactant

Cl₂ is the limiting reactant because it gives fewer moles of PCl₃

5. Calculate the mass of PCl₃

`\text{ Mass of PCl$_(3)$} = \text{0.3999 mol PCl$_(3)$} * \frac{\text{137.33 g PCl}_(3)}{\text{1 mol PCl}_(3)} = \textbf{54.92 g PCl}_\mathbf{{3}}`

Answer 2

`54.94~g~PCl_3`

Step-by-step explanation:

First we have to write the balanced reaction:

`2P~+~3Cl_2->~2PCl_3`

Then we have to find the limiting reagent. To do this we have to convert to moles first using the molar mass of each compound.

`P~=~30.97~g/mol`

`Cl_2~=~70.9~g/mol`

`12.39~g~P~(1~mol~P)/(30.97~g~P)=~0.4~mol~P`

`42.54~g~Cl_2~(1~mol~Cl_2)/(70.9~g~Cl_2)=~0.6~mol~Cl_2`

Then we have to divide by the amount of moles of each reagent in the balance reaction:

`(0.4~mol)/(2) =0.2`

`(0.6~mol)/(3) =0.2`

We have the same values for both compounds, so we can work with either of them.

Now, we can calculate the amount of
`PCl_3` produced:

`0.4~mol~P~(2~mol~PCl_3)/(2~mol~P)~(137.33~g~PCl_3)/(1~mol~PCl_3) ~=54.94~g~PCl_3`