Question

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?

Answer 1

`"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`