Respuesta :
Answer: 35.4 g
Explanation:
The balanced reaction is :
[tex]2P+3Cl_2\rightarrow 2PCl_3[/tex]
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of phosphorous}=\frac{15.5g}{31g/mol}=0.50moles[/tex]
[tex]\text{Moles of phosphorous chloride}=\frac{50.9g}{137g/mol}=0.372moles[/tex]
[tex]2P+3Cl_2\rightarrow 2PCl_3[/tex]
According to stoichiometry :
2 moles of phosphorous chloride are produced by = 3 moles of [tex]Cl_2[/tex]
Thus 0.37 moles of phosphorous chloride are produced by=[tex]\frac{3}{2}\times 0.372=0.558moles[/tex] of [tex]Cl_2[/tex]
Mass of [tex]Cl_2=moles\times {\text {Molar mass}}=0.558moles\times 71g/mol=35.4g[/tex]
Thus 35.4 g of chlorine reacted with the phosphorus
The mass of chlorine used in reaction has been 35.4 g. Thus, option C is correct.
The balanced chemical equation for the reaction has been:
[tex]\rm 2\;P\;+\;3\;Cl_2\;\rightarrow\;2\;PCl_3[/tex]
From the balanced reaction, 2 moles of phosphorus forms, 2 moles of phosphorus chloride.
The moles of phosphorus used has been given as:
[tex]\rm Moles=\dfrac{mass}{molar\;mass}[/tex]
The moles of phosphorus in 15.5 grams has been:
[tex]\rm Moles\;P=\;\dfrac{15.5}{31}\\Moles\;P= 0.50\;mol[/tex]
The moles of phosphorus chloride in 50.9 grams has been:
[tex]\rm Moles\;PCl_3=\dfrac{50.9 }{137} \\Moles\;PCl_3=0.372\;mol[/tex]
Since, 2 moles of phosphorus forms 2 moles phosphorus chloride. From 0.50 moles phosphorus, 0.372 mol phosphorus chloride has been formed.
Thus, chloride has been the limiting reagent, and the moles of chloride used in reaction has been equivalent to phosphorus chloride formed.
[tex]\rm 2\;mol\;PCl_3=3\;mol\;Cl_2\\0.372\;mol\;PCl_3=\dfrac{3}{2}\;\times\; 0.372mol\;Cl_2\\0.372\;mol\;PCl_3=0.558\;mol\;Cl_2[/tex]
The moles of chlorine used has been 0.558 mol. The mass of chlorine has been given as:
[tex]\rm Mass=Moles\;\times\;Molar\;mass\\Mass\;Cl_2=0.558\;\times\;71\;g\\Mass\;Cl_2=35.4\;g[/tex]
The mass of chlorine used in reaction has been 35.4 g. Thus, option C is correct.
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