Hess' Law for the enthalpy change of reaction of the decomposition of sodium bicarbonate

Question

I am currently writing a lab report that is about the Hess' law for the thermal decomposition of sodium hydrogen carbonate to for sodium carbonate. We did this by reacting sodium hydrogen carbonate with HCl as well as sodium carbonate with HCl. We found the values for those reactions using q=mcΔT to find the heat energy, then used ΔH=q/mol to find the enthalpy change. Then we applied Hess' law (simultaneous equations) to find the enthalpy change of reaction of decomposition. However, the value I got so far is +10.17, which is quite far off from the literature value from what I've heard. Please explain what I did wrong. Thank you so much!

My working out:

NaHCO3 - I used 25ml HCl and 1g of sodium bicarbonate, temperature change was -2.5

q=mcΔT

q= (25)(4.18)(-2.5)

q= -0.26125kJ

ΔH=q/mol

ΔH= -0.26125/0.011904

ΔH= -21.9468 kJ/mol

Na2CO3 - I used 25ml HCl and 1g of sodium carbonate, temperature change was 2.9

q=mcΔT

q= (25)(4.18)(2.9)

q= 0.30305kJ

ΔH=q/mol

ΔH= 0.30305/0.009435

ΔH= 32.1199 kJ/mol

Hess' law -

1. NaHCO3 + HCl --> NaCl + H2O + CO2 ΔH = -21.95
2. Na2CO3 + 2HCl --> NaCl + H2O + CO2 ΔH = 32.12 (flipped)

equation 1 - equation 2 = enthalpy of decomposition

-21.95-(-32.12) = 10.17 kJ/mol

Answer 1

The thermal decomposition of sodium bicarbonate would be

2NaHCO3 ==> Na2CO3 + CO2 + H2O

1.) NaHCO3 + HCl ==> NaCl + H2O + CO2 ... ∆H = -21.95

2.) Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2 ... ∆H = 32.12

Multiply (1) by 2: 2NaHCO3 + 2HCl ==> 2NaCl + 2H2O + 2CO2 ... ∆H = -43.9

Reverse (2): NaCl + H2O + CO2 ==> Na2CO3 + 2HCl ... ∆H = -32.12

Add them____________________________________________________________

2NaHCO3 + 2HCl + 2NaCl + H2O + CO2 ==> 2NaCl + 2H2O + 2CO2 + Na2CO3 + 2HCl

Final equation: 2NaHCO₃ ==> Na₂CO₃ + CO₂ + H₂O ∆H = -43.9 + - 32.12 = -76.02 kJ