Question

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?

Answer 1

ΔH=15000 J = 15KJ

Step-by-step 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