Consider the following reaction.
2 HCl(aq) + Ba(OH)2(aq)
Calculate the heat when 100.8 mL of 0.500 M HCl is mixed with 300.0 mL of 0.450 M Ba(OH)2. Assuming that the temperature of both solutions was initially 25.0°C and that the final mixture has a mass of 400.8 g and a specific heat capacity of 4.18 J/°C · g, calculate the final temperature of the mixture.
The Attempt at a Solution
using c=m/V and that n = m/M, I got that HCl is the Limiting Reactant and that there are 0.001382 moles of it. Then Q = delta H = -118*0.001382/2 = -0.081538
Plugging this into Q=mc(delta)T:
-0.081538 = 400.8 x 4.18 x (T_f - 25)
I got T_f = 24.99..., which is wrong.
Where did I go wrong? My assignment is due soon, so please help!