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.

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codiepienagoya codiepienagoya · Answer 1 · 2020-10-27T11:09:18+01:00

Answer:

The answer is "52.18 g"

Explanation:

Phosphorus mass = 12.39

molar mass of [tex]P_4 = 4 \times 30.974[/tex]

[tex]= 123.896 \ \frac{g}{mol}[/tex]

Phosphorus moles =[tex]\frac{mass}{molar \ mass}[/tex]

[tex]= \frac{12.39}{123.896} \\\\ = 0.10[/tex]

mass = 40.75 g

[tex]Cl_2[/tex] molar mass[tex]= 2 \times 35.45 \\\\[/tex]

[tex]= 70.9 \ \frac{g}{mol}[/tex]

[tex]Cl_2[/tex] moles = [tex]\frac{mass}{molar \ mass}[/tex]

[tex]= \frac{40.75}{70.9} \\\\ =0.57[/tex]

The balanced equation is: [tex]P_4 + 6 Cl_2 \longrightarrow 4 PCl_3[/tex]

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:

[tex]1 \ mole \ of \ P_4 \to 6 \ moles \ Cl_2\\\\0.1 \ mole \ P_4 = ?[/tex]

Moles of [tex]Cl_2[/tex] required = 0.6 moles

But only [tex]0.57[/tex], [tex]Cl_2[/tex] moles are less than the amount needed Thus, [tex]Cl_2[/tex] is reactant restricted

In the step2:

Find [tex]PCl_3[/tex]Theoretical performance [tex]PCl_3[/tex] the amount formed when the reactant is fully restricted

From equilibrium

[tex]6 \ moles\ Cl_ 2 \to 4 \ moles \ PCl_3 \\\\0.57\ moles \ Cl_2 = ?[/tex]

[tex]PCl_3[/tex] shaped mass = [tex]0.57 \times \frac{4}{6}[/tex]

[tex]= 0.38 \ moles[/tex]

[tex]PCl_3[/tex] shaped mass =[tex]moles \times molar\ mass[/tex]

Mass shaped by [tex]PCl_3[/tex] = [tex]0.38 \times 137.324[/tex]

[tex]= 52.18 \ g[/tex]