When1.42g of iron reacts with 1.80g of chlorine, 3.22g of FeCl2(s) and 8.60kJ of heat is produced.What is the enthalpy change for the reaction when 1 mole of FeCl2(s) is produced?

this is an enthalpy change problem;
the answer is ΔH= −338 kJ
I am not sure how to set up the problem to get this answer

1 Answer
zhirou
Jan 13, 2018

#-338kJ#.

Explanation:

Starting off with a balanced chemical equation...
#Fe + Cl_2 -> FeCl_2#

The question tells us that #"1.42 g"# of iron reacts with #"1.80 g"# of chlorine, we don't actually need this information.

The only thing we really need is the amount of iron(II) chloride produced, because that's what #ΔH# is based on for this question.

#"3.22 g"# of iron(II) chloride is produced, with a #ΔH# of #"8.60 kJ"#.

#"3.22 g"# of iron(II) chloride = #"0.0254 moles"#.
We get this from:
#"Moles in 3.22 g" = "3.22 g"/"mass of one mole of iron(II) chloride"#
#= 3.22/(55.85 + (2*35.45)) = "0.0254"#.

#∴# #"0.0254 moles"# of iron(II) chloride has a #ΔH# of #"-8.60 kJ"#

#∴# #"1 mole"# of iron(ii) chloride has a #ΔH# of #(-8.60)/0.02540434 = "-338 kJ"#

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