Show your work for all calculations. 1. When 1.104 grams of iron metal are mixed

Question

Show your work for all calculations.

1. When 1.104 grams of iron metal are mixed with 26.023 grams of hydrochloric acid in a coffee cup calorimeter, the temperature rises from 25.2 °C to a maximum of 33.5 °C. The reaction that occurs is given below.

2Fe(s)+6HCl(aq) 2FeCl3 (aq)+3H2 (g)

a) Determine the amount of heat (in J) absorbed by the reaction mixture. Assume that the specific heat capacity of the mixture is the same as the specific heat capacity of water.

b) How much heat (in J) was released by the reaction that occurred?

c) Is this reaction exothermic or endothermic? Is Hreaction positive or negative?

d) Under constant pressure conditions (as used in this experiment), the heat released by the reaction equals the reaction enthalpy, qreleased = Hreaction. Determine Hreaction in Joules per gram of metal used (J/g).

e) Determine Hreaction in kilojoules per mole of metal used (kJ/mol).

f) Determine Hreaction in kilojoules per mole for the balanced reaction equation provided (kJ/mol).

2. Consider the following three reactions:

2Fe(s)+6HCl(aq) 2FeCl3 (aq)+3H2 (g) HA

Fe2O3 (s)+6HCl(aq) 2FeCl3 (aq)+3H2O(l) HB

2H2 (g) + O2 (g) 2H2O(l) HC

Show how these equations must be summed together according to Hess’s Law to determine H for the combustion of iron (target equation shown below). Also show clearly how the H values of each of the three reactions must be manipulated to determine the enthalpy of combustion of iron.

4Fe(s)+3O2 (g) 2Fe2O3 (s) H=?

3. Using tabulated Hf° values in the text, determine the enthalpy change (in kJ/mol) that occurs during the formation of water from its elements as described by the equation:

2H2 (g) + O2 (g) 2H2O(l) H=?

Explanation / Answer

1. A

When 2 different substances that are initially at different temperatures come into contact an energy exchange will begin between them, the heat gained by one of the substances must be equal to the heat lost by the other.

We need to calculate the heat gained (or lost) by the calorimeter, we do this with the equation

q = m * cp * (Tf - Ti)

where q is the heat

m is the mass of the solution (26.023 + 1.104= 27.127grams)

cp is the specific capacity , 4.184 J/g C for water (also for the solution according to the statement)

T f and Ti are final and initial temperatures so

q calorimeter = 27.127 g * 4.184 J / g C * (33.5 - 25.2) = 942 Joules,

B) the calorimeter increased the temperature so the reaction released heat then the q rxn is -942 Joules, the negative sign denotes heat release

C) since the reaction released heat then we say that it is exothermic and enthalpy H rxn = q rxn = -942 Joules

D) to calculate the joules per gram of metal used we simply divide the heat of rxn by the grams of iron (1.104 grams)

H rxn = -942 J / 1.104 grams = -853.26 J/g

E) turn the joules from C to kilojoules, we do this dividing by 1000 then

heat of rxn = -0.942 KJ

find the moles of iron (molar mass of iron = 55.84)

moles = mass / molar mass = 1.104 / 55.84 = 0.019769 moles then

heat of rxn (KJ / mole) = -0.942 / 0.019769 = -47.65 KJ / mole

f) by looking at the equation we can see there are 2 moles of Fe that are reacting to get the product so to get the enthalpy of this reaction we just need to multiply the heat of rxn by 2

-47.65 * 2 = -95.3 KJ

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