Question

During an experiment, a student adds 0.339 g of calcium metal to 100.0 mL of 2.05 M HCl. 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, ΔHrxn.

Answer 1

moles of Ca = 0.339 / 40 = 8.475 x 10^-3

moles of HCl = 100 x 2.05 / 1000 = 0.205

mass of solution = 100 x 1 = 100 g

specific heat = 4.184 J / g oC

Q = m Cp dT

    = 100 x 4.184 x 11

Q = 4602.4 J

Ca (s)    +    2 HCl    ---------------> CaCl2 + H2 (g)

1                       2

8.475 x 10^-3    0.205

ΔHrxn. = - Q / n

           = - 4602.4 x 10^-3 / 8.475 x 10^-3

ΔHrxn = - 543 kJ/mol