Question

I drop an unknown quantity of Na metal in a coffee cup calorimeter that contains 150.0g if water with T initial = 23.66 Celcius. After the reaction with the metal, the temperature of the resulting solution is 34.81 Celcius. How much metal was dropped in the water for the reaction? Assume the mass of Na added to the is <<<< mass of the water used = 150.g. Assume Cw = C solution = 4.180 J/GC

Answer 1

Let the mass of the metal = m

heat absorbed by the water = mass * specific heat * change in temp

                                        = (m + 150) * 4.18 J/goC * (34.81-23.66)oC

As mass of the metal <<< to mass of water m+ 150 = 150

heat absorbed = 196.61 J

the reaction between Sodium and water is : 2Na + 2H2O = 2NaOH + H2

Enthalpy change = 368.6 kJ/mol (consult your textbook for this standard value)

amount of heat liberated in the coffee cup calorimeter = 196.61 J

moles of sodium reacted = 196.61 J/368.6 J/mol = 0.53 *10^-3 moles

mass of Na = 23 gm/mol * 0.53 *10^-3 moles = 0.012 gm

Amount of Na metal dropped = 0.012 gm