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## Homework Statement

Consider the following reaction.

2 HCl(

*aq*) + Ba(OH)2(

*aq*)

*aq*) + 2 H2O(

*l*) Δ

*H*= -118 kJ

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.

## Homework Equations

Q=mc(delta)T

c=m/V

## 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!