14.7 g of solid sodium hydrogen carbonate (MM = 84.008 g mol^-1) was reacted with 120 mL of 1.50 mol L^-1 hydrochloric acid solution (density 1.02 g mL^-1) in a calorimeter - HSC - SSCE Chemistry - Question 36 - 2024 - Paper 1

Using the concentration and volume of hydrochloric acid solution:

$n(HCl) = 1.50\, \text{mol L}^{-1} \times 0.120\, \text{L} = 0.180\, \text{mol}$

The reaction is 1:1, therefore hydrochloric acid is in excess and sodium hydrogen carbonate is the limiting reagent.

Using density to find the mass:

$\text{mass of solution} = (120\, \text{mL} \times 1.02\, \text{g mL}^{-1}) + 14.7\, \text{g} = 129\, \text{g}$

Using the temperature change and specific heat capacity:

$q = mc\Delta T = 129\, \text{g} \times 3.80\, \text{J K}^{-1} g^{-1} \times (11.5 - 21.5) = -4971\, ext{J} = -4.971\, ext{kJ}$

Finally, convert the heat released to per mole of sodium hydrogen carbonate:

$\Delta H = \frac{-4.92\, \text{kJ}}{0.175\, \text{mol}} = -28.1\, \text{kJ mol}^{-1}$