What is/are the purpose (s) of the styrofoam cups in this experiment? What are s

Question

What is/are the purpose (s) of the styrofoam cups in this experiment? What are some assumptions that we make during this experiment in order to calculate the of formation of ammonium sulfate? calculate the mass of the solution used if a student added 35.569 grams of ammonium sulfate to enough water to give 80.0 mL of solution The density the solution was 1.00 g/mL How many significant figures should your answer have? For the neutralization reaction (ie, Part I of this experiment) calculate the number of moles of the limiting reactant when 25.0 mL of 2.085 M H_2SI_4 (aq) is mixed with 50.0 mL of 1.873 M NH, (aq) using the molar ratio of the reaction. Show your work with the correct units and number of significant figures If a student calculated the number moles of limiting reactant for the above neutralization reaction (See Question 4 above) to 0.00568 moles of ammonium sulfate and the heat (ie., Q) of the reaction to be 680 J calculate the enthalpy change of this reaction. Show your work with the units and number of significant figures.

Explanation / Answer

1. A calorimetric experiment involve precise determination/measurement of the heat changes inside the system. Therefore, heat loss or gain with the surrounding air or any other medium needs to be minimized. The more these other heat exchanges are reduced, the more true the mathematical equation will be. A cup made up of Styrofoam makes a good adiabatic wall (i.e. zero heat loss or gain with the surrounding air) and helps keep all the heat released or absorbed by the reaction inside the cup so we can measure it.

2. To simplify the calculations in any calorimetric experiments, we generally make three assumptions:

3. Given quantities: Mass of ammonium sulfate used = 35.569 g, Total Volume of the solution = 80 mL, Density of the soultion = 1.00 g/mL.

1.00 g/mL = Total mass of the solution (g) / 80 mL

=> Total mass of the solution (g) = (1.00 g/ ml) x (80 mL) = 80 g

Total mass of the solution = Mass o the Ammonium sulfate + Mass of the solution used

80.000 g = 35.569 g + Mass of solution used (g)

=> Mass of soultion used = 80.000 - 35.569 = 44.431 g

Note: The mass is calcuated upto Three significant figures because the traditional weighing balances are accurate upto three significant figures (example: see mass of the mentioned Ammonium sulfate)

4. The balanced acid-base neutralization reaction is:

2NH3 + H2SO4 -> (NH4)2SO4

and then based on the number of moles of acid and base, we need to decide which is going to be a limiting reactant

Step 1. Volume conversion of acid and base from mL to Litre:

25.0 mL of acid (H2SO4) = 25 / 1000 L of H2SO4= 0.025 L of H2SO4

50.0 mL of base (NH3) = 50 / 1000 L of NH3 = 0.050 L of NH3

Step 2. Number of moles calculation using above Molarity expression and above calculated volumes and given Molarities:

Molarity (moles/Litre) = No. of moles (n) / Volume of solution (in L)

For H2SO4,

2.085 M = Number of Moles / 0.025

=> No. of moles of H2SO4 = 2.085 x 0.025 = 0.0521

For NH3,

1.873 M = Number of Moles / 0.050

=> No. of moles of NH3 = 1.873 x 0.050 = 0.0937

Therefore, the using below balanced chemical reaction:

2NH3 + H2SO4 -> (NH4)2SO4

and above calculated number of moles, it is clear that the twice the number of moles of NH3(2 x 0.0937 = 0.1874) is far high than the number of moles of H2SO4 (0.0521).

Therefore, in present case H2SO4 acts as a limiting reactant with number of moles equals to 0.0521.

Note: Corresondingly the number of moles of limiting product ((NH4)2SO4) using banced chemical reaction will be also 0.0521.

5. The standard enthaly chnge is always measured as amount of heat released or gained per mole of the substance.

Here, 0.00568 moles of ammonium sulfate is formed with the absorbance of 5680 J of energy (Note: no negative sign is given in the question). Therefore, using:

Enthalpy change = Amount of energy released or absorbed / Number of moles of Ammonium Sulfate

Enthapy change (J / mol) = (5680 J) / 0.00568

= 1,000,000 J/ mol

However, enthalpy chage is measured in Kilojoule / mole. Therefore, in present case, the enthalpy change is:

Enthapy change (KJ / mol) 1,000,000 J/ 1000 = 1000 KJ/ mol

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