A 2.500-mol sample of PCl5 dissociates at 160oC and 1.00 atm to give 0.338 mol of phosphorus trichloride, PCl3, at equilibrium. PCl5(g) <---> PCl3 + Cl2(g) ?

What is the number of moles of PCl5(g) in the final reaction mixture? Type the number.

1 Answer
Truong-Son N.
Mar 18, 2018

If you write out the ICE table, this is readily found.

#"PCl"_5(g) rightleftharpoons "PCl"_3(g) + "Cl"_2(g)#

#"I"" "2.500" "" "" "0" "" "" "0#
#"C"" "-x" "" "" "+x" "" "+x#
#"E"" "2.500-x" "x" "" "" "x#

where the terms in the ICE table are mols initially, or the change in mols, or the mols at equilibrium.

But we know that #x = "0.338 mols"#. So what are the mols of #"PCl"_5(g)# that are left? Report your answer to three decimal places.

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