[daisychain]

A 100 cm3 of 2 mol dm-3 hydrochloric acid was poured into a polystyrene cup. 8 grams of anhydrous sodium carbonate was added with.

Initial temperature was 20.0 C and final temperature was 25.0 C.

The balanced equation I got was 2HCl (aq) + Na2CO3(aq) → 2NaCl (aq) + CO2(g) + H2O(l)

How do I got about calculating the enthalpy of the reaction in kJ per mole?

Also what type of reaction is it?

Thanks!

[Borek]

Per mole of what?

Regardless of what the answer is, start calculating number of moles.

[daisychain]

I think I've calculated the enthalpy right.. would anyone mind checking it? 

(i) number of moles of HCL
1000cm3 HCL solution contain 1mol of HCL
Number of moles of HCL= MV/1000
M= molarity of the solution (mol dm-3)
V= volume of the solution (cm3)
(2 x 100)/1000= 0.2 mol

(ii) number of moles of Na2CO3
n= m/M
m= given mass (g)
M= molar mass(g mol-1)

M= (2 x 22.99) +(1 x 12.01) + (3 x 16.00)= 105.99 g mol-1

N= 8/105.99= 0.075 moles (3 s.f)

The ratio is 2 :1 so which means Na2CO3 is the limiting reactant.

(iii) Delta temperature
TFINAL-TINITIAL
25.0 ˚C - 20.0 ˚C= 5.0˚C (±1 ˚C)

(iv) Heat released during reaction
Q= mc ΔT

m= 100 cm3= 100 ml
Density of water = 1g mol
So in 100ml, its 1g mol x100ml = 100g

(100g)(4.18 J/K.C)(5.0˚C)= 2090 Joules
Q= 2.09 kJ

ΔH= Q/mols
ΔH= 2.09/(0.075 )= 27.87 kJ/ mol Na2CO3

um is this also an endothermic reaction?

[Borek]

um is this also an endothermic reaction?

Temperature went up - so the reaction was producing heat. Is it exo, or endo?