In the reaction between formic acid (HCHO2) and sodium hydroxide, water and sodium formate (NaCHO2) are formed. To determine the heat of reaction, 75.0 mL of 1.07 M HCHO2 was placed in a coffee cup calorimeter at a temperature of 21.9 °C, and 45.0 mL of 1.78 M NaOH, also at 21.9 °C, was added. The mixture was stirred quickly with a thermometer, and its temperature rose to 25.5 °C. Write the balanced chemical equation for the reaction. Calculate the heat of reaction in joules. Assume that the specific heats of all solutions are 4.18 J g-1°C-1 and that all densities are 1.00 g mL-1. What is the heat of reaction per mole of acid (in units of kJ mol-1).
Write the balanced reaction that takes place:
Heat of reaction = ________________---- kJ/mol

To balance the equation, the elements must be in the same amount in the reactants and the products. As we can see, there are the same amount of elements on each side, so the equation is balanced.

The heat flows can be calculated by:

Q = m*cp*ΔT

Where m is the mass of the substances (mass of formic acid + mass of sodium hydroxide), cp is the specif heat, and ΔT is the variation at temperature (final - initial).

Mass is the density multiplied by the volume, so:

m = 1*75 + 1*45 = 120 g

Q = 120*4.18*(25.5 - 21.9)

Q = 1,805.76 J = 1.81 kJ

The number of moles of the reactants can be calculated by the volume multipled by the concentration:

HCHO₂ = 0.075L * 1.07 mol/L = 0.08025 mol

NaOH = 0.045L * 1.78 mol/L = 0.0801 mol

So, NaOH is limiting (stoichiometry is 1:1, so it's necessary the same amount of the reactants), and the heat of the reaction will be calculated by it.

ΔH = Q/n

ΔH = 1.81/0.0801

ΔH = 22.6 kJ/mol