H2a and boh are acid and base and they react according to the following balanced equation: h2a(aq) + 2 boh(aq) → b2a(aq) + 2 h2o(l) if 0.20 moles of boh reacted with excess amount of h2a solution and 1500. j of heat was released by the reaction, what is the ∆hrxn for the reaction as written above?

ΔH=15000 J  =  15KJ

Explanation:

In this exercise  you have find the enthalpy of reaction this is the difference between enthalpy of reactans and products,

For the following equation

H2A(aq) + 2 BOH(aq) → B2A(aq) + 2 H2O(l)

We know that 0.20 moles of BOH reacted with excess amount of H2A solution and 1500. J

so,

(2mol/0,2mol)*1500J=15000J

for de reactions exothermics tha enthalpy is negative so:

ΔH=15000 J  =  15KJ

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hope i !