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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 20.8 °C, and 45.0 mL of 1.78 M NaOH, also at 20.8 °C, was added. The mixture was stirred quickly with a thermometer, and its temperature rose to 25.3 °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).

• chemistry -

NaOH + HCOOH ==> H2O + HCOONa

q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
[note that mass H2O = mL HCOOH + mL NaOH since the density of H2O is 1.0 g/mL.]

delta H rxn/gram acid = q/grams HCOOH
dH rxn/mol acid = (q/g HCOOH)*molar mass HCOOH.

• chemistry -

6.69 joules

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