Answer :
Answer:
[tex]54.94~g~PCl_3[/tex]
Explanation:
First we have to write the balanced reaction:
[tex]2P~+~3Cl_2->~2PCl_3[/tex]
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.
[tex]P~=~30.97~g/mol[/tex]
[tex]Cl_2~=~70.9~g/mol[/tex]
[tex]12.39~g~P~\frac{1~mol~P}{30.97~g~P}=~0.4~mol~P[/tex]
[tex]42.54~g~Cl_2~\frac{1~mol~Cl_2}{70.9~g~Cl_2}=~0.6~mol~Cl_2[/tex]
Then we have to divide by the amount of moles of each reagent in the balance reaction:
[tex]\frac{0.4~mol}{2} =0.2[/tex]
[tex]\frac{0.6~mol}{3} =0.2[/tex]
We have the same values for both compounds, so we can work with either of them.
Now, we can calculate the amount of [tex]PCl_3[/tex] produced:
[tex]0.4~mol~P~\frac{2~mol~PCl_3}{2~mol~P}~\frac{137.33~g~PCl_3}{1~mol~PCl_3} ~=54.94~g~PCl_3[/tex]
Answer:
[tex]\text{54.92 g PCl}_{3}}[/tex]
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
[tex]\text{Moles of P} = \text{12.39 g P} \times \dfrac{\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} \times \dfrac{\text{1 mol Cl$_{2}$}}{\text{70.91 g Cl$_{2}$}} = \text{0.5999 mol Cl$_{2}$}[/tex]
3. Calculate the moles of PCl₃ from each reactant
[tex]\textbf{From P:}\\\text{Moles of PCl$_{3}$} = \text{0.4006 mol P} \times \dfrac{\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$_{2}$} \times \dfrac{\text{2 mol PCl$_{3}$}}{\text{3 mol Cl$_{2}$}} = \text{0.3999 mol PCl$_{3}$}[/tex]
4. Identify the limiting reactant
Cl₂ is the limiting reactant because it gives fewer moles of PCl₃
5. Calculate the mass of PCl₃
[tex]\text{ Mass of PCl$_{3}$} = \text{0.3999 mol PCl$_{3}$} \times \dfrac{\text{137.33 g PCl}_{3}}{\text{1 mol PCl}_{3}} = \textbf{54.92 g PCl}_\mathbf{{3}}[/tex]