Question

0.15g Mg reacted with 50ml 1M HCL in a calorimeter: The calorimeter constant is 9.3

Mg (s) + 2HCL (aq) ---> Mg2+ (aq) + 2Cl- (aq) + H2 (g)

Mass of the empty calorimeter (g)	18.600g
Initial temperature of the calorimeter (°C)	21.5 C
Maximum temperature in the calorimeter from the reaction (°C)	34.5C
Calculate delta T by subtracting (b) from (c) (°C) delta T = Tmaximum - Tinitial	13.0C
Mass of the calorimeter and its contents after the reaction (g)	68.738g
Calculate the mass of the contents of the calorimeter (g) by subtracting (a) from (e)	50.138g
Calculate the moles of Mg reacted (MW = 24.305 g/mole)	0.006mol

1) Calculate the heat released into the solution for the reaction, according to:

Q = (C_cal * deltaT) + (mass of water * C_p(water) * deltaT)

2) Find the molar heat of reaction for each experiment in units of Kilo-Joules / (mole of Mg) by dividing the heat of reaction by the moles of Mg used. Show all work and include units for all values.

3) Using your calculated value for molar heat of reaction, how much heat would be given off in kilojoules if 3.0 grams of magnesium were to react with excess hydrochloric acid. Show work for your calculation.

*****I need help with setting up the calculations for each of these*****

Answer 1

Ans. Amount of heat gained or lost by a substance is given by-

            q = m s dT                            - equation 1

Where,

q = heat gained or lost

m = mass

s = specific heat

dT = Final temperature – Initial temperature

#1. Step 1: Mass of solution/ water = 50.0 mL (= mass of HCl).

It’s assumed that the density and specific heat of solution is the same as that of water. Also, the mass of solid Mg is negligible when compared to that of solution.

Heat gained by water, q1 = 50.0 g x (4.184 J g^-10C-1) x 13.0⁰C

            Hence, q1 = 2719.6 J

#Step 2: Heat gained by Calorimeter, q2 = m x C

            Where, C = Calorimeter constant

            Or, q2 = 18.6 g x (9.30 J ⁰C^-1)

            Hence, q2 = 172.98 J

#2. Step 3: During establishment of thermal equilibrium, the total heat lost by Mg-HCl reaction is gained by the calorimeter and solution.

So,

Heat lost during Mg-HCl reaction = - (Heat gained by water + Heat gained by Calorimeter)

Or, Heat lost during Mg-HCl reaction = - (2719.6 J + 172.98 J) = -2892.58 J

Hence, Heat lost during Mg-HCl reaction = -2892.58 J

Note: The –ve sign of (- 2892.58 J) indicates that heat is being released during the reaction.

#Step 4. Moles of Mg reacted = Mass / Molar mass

                                                = 0.15 g / (24.305 g/ mol)

                                                = 0.006172 mol

Now,

Heat released per mol Mg reacted with HCl = Heat released / Moles of Mg reacted

                                                = (-2892.58 J) / 0.006172 mol

                                                = -468661.70 J/ mol

                                                = -468.88 kJ/ mol

Hence, molar enthalpy of Mg-HCl reaction = -468.88 kJ/ mol

#3. Step 5: Moles of Mg = 3.0 g / (24.305 g/ mol) = 0.1234 mol

Amount of heat released by 3.0 g Mg = Moles of Mg x Molar enthalpy of reaction’

                                                = 0.1234 mol x (-468.88 kJ/ mol)  

                                                = -57.86 kJ