For the reaction: H2(g)+1/2O2(g)=H2O(g). The #DeltaH_(rxn)^@# = -241.8 Jmol ^-1 What quantity of heat, in kj, evolved when a 72.0 g mixture containing equal parts of H2 and O2, by mass, is burned?

1 Answer
Truong-Son N.
Mar 5, 2018

Well, once you correct your #DeltaH_(rxn)^@#, you should get #-"544.1 kJ"#...

Well, the reaction is for the formation of #"1 mol"# of #"H"_2"O"(g)#, so it is due to the reaction of #"1 mol H"_2(g)# where #DeltaH_f("H"_2"O"(g))^@ = DeltaH_(rxn)^@#.

You'll need #"0.5 mols O"_2(g)# for every #"1 mol H"_2(g)#, so find the mols of each reactant.

#"36.0 g H"_2 xx ("1 mol")/("2.0158 g H"_2) = "17.9 mols H"_2#

#"36.0 g O"_2 xx ("1 mol")/("31.998 g O"_2) = "1.13 mols O"_2#

This is more than twice the #"mols"# stated in the reaction, so... #DeltaH_(rxn)^@# will be more than twice what is shown.

Also, there is much more than twice as many mols of #"H"_2# compared to #"O"_2#.

We can then see that #"O"_2# is very much the limiting reactant, so the reaction forms #"2.25 mols H"_2"O"(g)# due to the reaction of #"1.13 mols O"_2(g)# with #"2.25 mols H"_2(g)#.

Therefore:

#color(blue)(DeltaH_(rxn)^@) = -"241.8 kJ"/"mol" xx "2.25 mols H"_2"O" = color(blue)(-"544.1 kJ")#

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