Question

During an experiment, a student adds 0.339 g of calcium metal to 100.0 mL of 2.05 M HCI. The student observes a temperature increase of 11.0 °C for the solution. Assuming the solution's final volume is 100.0 mL, the density is 1.00 g/mL, and the specific heat is 4.184 J/(g-C), calculate the heat of the reaction, A Hn Ca(s) 2H* (aq)Ca2 (aq) +H,(g) kJ/mol AH =

Answer 1

We have relation, density = Mass / Volume

$\therefore$ Mass = density $\times$ volume

$\therefore$ Mass of solution = 1.00 g/ml $\times$ 100.0 ml = 100 g

Specific heat capacity of solution ( C) = 4.184 J / g ⁰ C

Change in temperature ( T) =11.0 ⁰ C

In this case, q reaction = - q solution

Hence , Enthalpy change of solution =- [m x C x   T]

= - [ 100 g x 4.184 J / g ⁰ C x (11 ⁰ C )

= - 4602 J

Moles of Ca = Mass / Molar mass = 0.339 g / 40.08 g/mol = 0.00846 mol

Moles of HCl = Molarity $\times$ volume of solution in L

Moles of HCl = 2.05 mol / L $\times$ 0.1000 L = 0.205 mol

from the moles of Ca and HCl, it is clear that Ca is limiting reactant.

Hence, enthalpy change of reaction will be =expressed in terms of Ca.

For the reaction of 0.00846 mol Ca , q = - 4602 J .

Therefore, heat produced when 1 mole of Ca reacts = 1 mol Ca ( - 4602 J / 0.00846 mol ) = - 544 kJ

ANSWER : - 544 kJ / mol