Enthalpy Change of Formation

I'm going through the questions in my nuffield textbook to make notes to revise from and I just want you to check whether my answer to tis question is correct =S I'm not sure about one thing.

The Question:

The enthalpy change for the decomposition of Calcium Carbonate

Unparseable latex formula:

\mathrm{CaCO_3 _{(s)} \to CaO_{(s)} + CO_2_{(g)}}

is impossible to determine directly so the following method was devised:

Two pieces of calcium carbonate were chosen, each with mass 1.25g. One piece, A, was put into 20cm^3 of dilute hydrochloric acid, the temperature of which rose by 2°C. The other piece was heated in the hottest flame available for ten minutes to decompose it to Calcium Oxide. It was then allowed to cool to room temperature before it was added to 20cm^3 (an excess) of dilute hydrocholric acid. The temperature of the acid rose by 12°C.
(Molar Mass of Calcium Carbonate = 100gmol^-1.)

a) (i) Calcuate the energy produced by the reaction of each solid sample with the acid. Use the relationship:
Energy Produced (J) = 4.18 x mass of solution x temperature rise

4.18 \times 20 \times 2 = 167.2 \mathrm{J}

4.18 \times 20 \times 12 = 1003.2 \mathrm{J}

Here, I'm not sure whether I should have added the 1.25 onto the the 20cm^3 and assumed the density of the solution was 1gcm^3. Is this right?

(ii) State two assumptions that have to be made when using the relationship in (i).

That the specific heat capacity of the solution is 4.18J g^-1 K^-1 and that the reaction was under standard conditions.

(iii) How many moles of CaCO3 were there in each of the original pieces?

1.25/100 = 0.0125 moles

(iv) Use your answers to a (i) and a (ii) to calculate enthalpy changes for the two reactions with the acid,

\Delta H_A

(from CaCO3) and

\Delta H_B

(from CaO). Your answers should be given to 2SF and include signs and units.

1)

\mathrm{CaCO_3 + 2HCl \to CaCl_2 + H_2O + CO_2}

\mathrm{CaCO_3 \to CaO + CO_2}

\mathrm{CaO + 2HCl \to CaCl_2 + H_2O}

0.0125 moles of CaCO3
HCl 2 : 1 CaCO3
Therefore we have 0.025 moles of HCl.

\frac{167.2}{0.025} = \mathrm{6688 Jmol^{-1}}

\Delta H_A = -6.7 \mathrm{kJ mol^{-1}}

\frac{1003.2}{0.025} = 40128 \mathrm{Jmol^{-1}}

\Delta H_B = -40 \mathrm{kJmol^{1}}

b) *Hess Cycle has been drawn, with arrows pointing DOWNWARDS from the reactants t o the elements and DOWNWARDS fromthe products to the elements box.*

Calculate the vaue of

\Delta H_{reaction}

\Delta H_{reaction}

\Delta H_B

\Delta H_A

\Delta H_{reaction}

= -6.7 + 40

= +33.3 kJmol^{-1}

The data book says the answer is +178.3kJ mol^-1, which is why I think it's totally wrong :s-smilie:

. Surely the way the experiment was conducted couldn't affect the results this much? Can someone explain where I've went wrong?
Sorry for the long post =p