When 1.50g of Ba(s) is added to 100.0g of water in a container open to the atmosphere, the reaction shown below occurs and the temperature of the resulting solution rises from 22.00 celsius to 33.10 celsius. If the specific heat of the solution is 4.18J/g*celsius, calculate the change of enthalpy for the reaction, as written below.
Ba(s) + 2H2O(l) -> Ba(OH)2(aq) + H2(g)
q = ms(delta)T
q(surrounding) = -q(system)
(delta)H = q (constant pressure)
The Attempt at a Solution
First, I calculated the limiting reagent of the solution (Ba(OH)2) using both the masses of Ba and H2O. The limiting reagent happened to be .0109 moles of Barium Hydroxide (also the same number of moles for solid Barium). After that, I multiplied the moles by the molar mass of Barium Hydroxide, which gave me the mass of 1.87g. After I found delta T, I used the following information to find the heat of the surrounding by performing the following equation:
q(surrounding) = (1.87g)(4.18J/g*celsius)(11.10 celsius)
q = 86.8 J
Since the reaction is an exothermic reaction, the q will become negative, becoming -86.8J.
After that, I calculate the entire reaction by using the following equation:
[(.0109 mol Ba(OH)2) / -86.8J)] = [1 mol Ba(OH)2/x]
.0109x mol Ba(OH)2 = -86.8J * mol Ba(OH)2
x = -7963.30 J or -7.96 kJ
The correct answer happens to be -431 KJ/mol. I have absolutely no clue how it got there. I even tried to work backwards from the answer, but it still makes no sense. I have my final exam next week and I really need to understand this 100%.