Question

If possible please give the formula or steps required for me to solve these questions, as I would like to go through the steps myself rather than just receiving answers. Thanks.

Information

BACKGROUND:

This is part 1 of an enthalpy experiment involving:

Use of NaCl and HCl to determine the heat capacity of a coffee-cup (styrofoam cup) caloimeter;

And the following to determine enthalpy change in the reaction of Magnesium with Hydrochloric acid;

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

H+(aq) + HO-(aq) --> H2O(aq)

Change H = -57,320 J

DATA:

Mass of weighing paper: 0.408g

Mass of weighing paper & Mg: 0.907

Volume of HCl solution: 50.00mL

Molarity of HCl solution: 2.144M

Volume of H2O added to HCl solution: 100.00mL

Initial temperature of HCl solution: 22.0 deg. C

Final temperature of HCl solution: 41.0 deg. C

Peak temperature of HCl solution: 42.0 deg. C

Problems

1.) Mass of final NaCl solution assuming the density of a 1M NaCl solution is 1.04 g/mL?

2.) Heat actually produced from the HCl-NaOH reaction assuming that the heat capacity of a 1M NaCl solution is 3.93 J/g deg. C?

3.) Moles of water formed in the reaction?

4.) Total theoretical heat expected to be produced by forming the number of moles of water assuming that 57,320 J are produced per mole of water formed?

5.) Heat absorbed by the calorimeter?

6.) Heat capacity of the calorimeter?

Answer 1

Part 1.

Mass of final NaCl solution (m) = 1 mol/L * 58.5 g/mol = 58.5 g = 58.5 g/(1.04 g/mL) = 56.25 mL

Explanation: Density = mass/volume, molecular weight of NaCl = 58.5 g/mol, molarity = mol/L

Part 2.

The heat produced from the HCl-NaOH reaction (q) = m*C*$\Delta$T

Where 'C' is the specific heat of the solution = 3.93 J/g. ^oC

$\Delta$T is the change in temperature = final temperature - initial temperature = 41-22 = 19 ^oC

i.e. q = 58.5 g * 3.93 J/g. ^oC * 19 ^oC = 4368.195 or 4.368 kJ

Part 3.

Moles of water formed in the reaction = moles of NaCl = 1 mol

Part 4.

Since the moles of water = 1 mol

The amount of theoretical heat expected to be produced = 1 mol * 57320 J/mol = 57320 J or 57.32 kJ