Calculating Delta-H for a Reaction using a Coffee-Cup Calorimeter Method

[sp3sp2sp]

Homework Statement

The addition of hydrochloric acid to a silver nitrate solution precipitates silver chloride according to the reaction:

AgNO3(aq)+HCl(aq)→AgCl(s)+HNO3(aq)

When 500 mL of 0.100 MAgNO3 is combined with 500 mL of HCl in a coffee-cup calorimeter, the temperature changes from 23.40 ∘C to 24.21 ∘C. Calculate ΔHrxnfor the reaction as written. Use 1.00 g/mL as the density of the solution and C=4.18J/(g⋅∘C) as the specific heat capacity

Homework Equations

MM AgNO3 = 169.88g.mol
q = mcdeltaT = deltaH_rxn

The Attempt at a Solution

q = [(500mL/1000)(0.100M)(169.88g/mol) ] * 4.18J/g/degC * 13.31degC = 28.75J = q =ΔHrxn for 0.05mol
28.75J * 20 = 580J = 0.58kJ (2 sig-fig)
= wrong
Thans for any help

 

[Borek]

Please elaborate on

q = [(500mL/1000)(0.100M)(169.88g/mol) ] * 4.18J/g/degC * 13.31degC

None of the numbers here makes sense to me

 

[sp3sp2sp]

thanks for the reply. It is from q = m*c*deltaT. I did make mistake for delta T, which is corrected below.
First I calculated the grams of AgNO3:
[(500mL/1000)(0.100M) * (169.88g/mol) = 8.494g

I was provided C=4.18J/(g⋅∘C) as the specific heat capacity in question stem.

temp change is T-final - T-initial = 24.21 - 23.40 = 0.81degC

then q = [(500mL/1000)(0.100M)(169.88g/mol) ] * 4.18J/g/degC * 0.81degC = 28.76J

q = -enthalpy = -28.76J for .05mol of AgNO3.

20* .05mol = 1 mol of AgNO3

so 20 * -28.76J = -575.2J = -0.5752kJ

Answer needed to be to two sig figs, so = 0.58kJ

I know there's mistakes in this but I am not sure where they are...thanks again for any help

 

[Chestermiller]

You determined that 0.05 moles of AgNO3 reacted. This is correct.

You had 1000 ml of liquid that were heated 0.81 C. How many joules of heat does this correspond to? How many joules per mole of AgNO3 does this correspond to?