How many moles of pcl5 can be produced from 57.0 g of cl2 (and excess p4)?

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Question:

How many moles of pcl5 can be produced from 57.0 g of cl2 (and excess p4)?

Answers

The reaction will take place according to the equation;P4 + 10Cl2 = 4PCl51 mole of Cl2 contains 70.9 g/molTherefore, the number of moles of Cl2 will be;  = 57 g/ 70.9 g/mol  = 0.804 molesThe mole ratio of Cl2 : PCl5 = 10:4Therefore; the number of moles of PCl5 = (0.804/10)×4        = 0.3216 moles

Answer : The number of moles of [tex]PCl_5[/tex] produced can be 0.321 moles.Solution : Given,Mass of [tex]Cl_2[/tex] = 57.0 gMolar mass of [tex]Cl_2[/tex] = 71 g/moleFirst we have to calculate the moles of [tex]Cl_2[/tex].[tex]\text{ Moles of }Cl_2=\frac{\text{ Mass of }Cl_2}{\text{ Molar mass of }Cl_2}=\frac{57.0g}{71g/mole}=0.803moles[/tex]Now we have to calculate the moles of [tex]PCl_5[/tex]The balanced chemical reaction is,[tex]P_4+10Cl_2\rightarrow 4PCl_5[/tex]From the reaction, we conclude thatAs, 10 mole of [tex]Cl_2[/tex] react to give 4 mole of [tex]PCl_5[/tex]So, 0.803 moles of [tex]Cl_2[/tex] react to give [tex]\frac{0.803}{10}\times 4=0.321[/tex] moles of [tex]PCl_5[/tex]Therefore, the number of moles of [tex]PCl_5[/tex] produced can be 0.321 moles.

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