How much heat will be released when 12.0 g of #H_2# reacts with 76.0 g of #O_2# according to the following equation? #2H_2 + O_2 -> 2H_2O# #DeltaH#= -571.6 kJ?

1 Answer
anor277
Nov 19, 2016

#"Dioxygen gas"# is the limiting reagent. Over #1.3xx10^3*kJ# are evolved.

Explanation:

#2H_2(g) + O_2(g) rarr 2H_2O(l)# #DeltaH=-576.6*kJ*mol^-1.#

#"Moles of dihydrogen"=(12.0*g)/(2.016*g*mol^-1)=5.95*mol#.

#"Moles of dioxygen"=(76.0*g)/(32.0*g*mol^-1)=2.38*mol#.

Given the stoichiometry, clearly, there is a insufficient molar quantity of dioxygen for complete combustion. At most #4.75*mol# dihydrogen can react (i.e. #2xx2.38*mol)# according to the given equation.

And thus energy released can based on the molar quantity of #"dioxygen gas"# #=# #2.38*molxx-576.6*kJ*mol^-1=??kJ#

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