What is/are the purpose(s) of the Styrofoam^cups in this experiment? What are so

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 enthalpy 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 of the solution was 1.00 g/mL. How many significant figures should your answer have?_____ For the neutralization reaction (i.e., Part I of this experiment), calculate the number of moles of the limiting reactant when 25.0 mL of 2.085 M H_2SO_4 (aq) is mixed with 50.0 mL of 1.873 M NH_3 (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 above) to be 0.00568 moles of ammonium sulfate and the heat (i.e., Q) of the reaction to be 5680 J, calculate the enthalpy change of this reaction. Show your work with the correct units and number of significant figures.

Explanation / Answer

Styrofoam cups are thermal insulators

In this experiment if heat is released to the sorroundings then it will lead to miscalculations

so we are using styrofoam cups as this experiment involves release of heat.

2) this question you have to answer from your introduction section as mentioned in the question. as you didn't post it i could not answer this

3)Volume of solution = 80 ml

density of solution = 1 g/ml

Mass of solution = 80 ml * 1 g/ml = 80 g Answer

4) Given

H2SO4

volume = 25 ml = 0.025 L Molarity = 2.085 M (or mol/L)

No. of moles = volume * molarity = 0.025 L * 2.085 mol//L = 0.0521 moles

NH3

Volume = 50 ml = 0.05 L Molarity = 1.873 M (or mol/L )

No. of moles = 0.05 L * 1.873 mol/L = 0.09365 moles

reaction is

2 NH3 + H2SO4 ----> (NH4)2SO4

2 mole 1 mole 1 mole

so for 0.0521 moles of H2SO4 we need 2 * 0.0521 = 0.1042 moles of NH3 but we have only 0.0935 moles of NH3

so NH3 is the limiting reactant with 0.0935 moles

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