Question 9: calculate delta H rxn mass of MgO:2.8044 g initial temp:25 C final t
ID: 568748 • Letter: Q
Question
Question 9: calculate delta H rxn
mass of MgO:2.8044 g
initial temp:25 C
final tempt:46.34 C
1.000M HCl 100mL
Ture &t; tke el solution. Drag the weigh sample over to the calorimeter and drop the sample in. Observe the change in temperature until it reaches a maximum and then record data for an addi temperature window. (You can click on the clock on the wall labeled Accelerate to paper containing the Mg itional 20-30 seconds. Click Stop in the ) A data link icon will appear in the lab book. Click the data link icon and record the temperature before adding the MgO and the highest temperature after adding table. (Remember that the water will begi n to cool down after reaching the equilibrium temperature.) the MgO in the data Data Table MgO/HC Mass MgO initial temperature (C) final temperature (C) 4. Is the observed reaction endothermic or exothermic? What will be the sign of AHrm? 5. Calculate the change in temperature, AT. Record your results in the results table on the following page 6. Calculate the mass of the reaction mixture in the calorimeter. (To do this, assume that the density of the HCI solution originally in the calorimeter can be approximated with the density of water (1L.0 g/mL). Record your results in the results table. 91 Copyright © 2014 Pearson Education, Inc.Explanation / Answer
Ans. # Mass of HCl solution = Volume x Density = 100.0 mL x 1.000 g mL-1 = 100.0 g
Mass of reaction mixture = Mass of HCl + Mass of MgO
= 100.0 g + 2.8044 g
= 102.8044 g
#Step 1: Heat gained by a solution is given by-
q = m s dT - equation 1
Where,
q = heat gained
m = mass
s = specific heat
dT = Final temperature – Initial temperature
Assumption: Since actual specific heat of reaction mixture or HCl solution in NOT mentioned in the question, it’s assumed that the specific heat of reaction mixture is equal to that of water.
Putting the values in equation 1-
q = 102.8044 g x (4.184 J g-1 0C-1) x (46.34 – 25.0)0C
Or, q = 9179.0512 J
# The total amount of heat gained by the solution to increase its temperature must be equal to the total amount of heat released during the reaction between MgO and HCl.
So,
Heat released during reaction between MgO and HCl = - 9179.0512 J
Note: the –ve sign indicates that heat is being released during the reaction.
#Step 2: Determine limiting reagent:
# Balanced reaction: MgO(s) + 2 HCl(aq) ---------> MgCl2(aq) + H2O(l)
Stoichiometry: 1 mol MgO reacts with 2 mol HCl.
So, molar ratio of reactants = MgO : HCl = 1 : 2
# Moles of MgO = Mass / Molar mass = 2.8044 g / (40.3044 g/ mol) = 0.06958 mol
Moles of HCl = Molarity x Volume of solution in liters = 1.00 M x 0.100 L = 0.100 mol
Now,
Experimental molar ratio of reactants = Moles of MgO / Moles of HCl
= 0.06958 mol : 0.100 mol
= 1 : 1.4
# Comparing the theoretical and experimental molar ration, the experimental moles of HCl are less than its theoretical value 2 while keeping the moles of MgO constant. So, HCl is the limiting reactant.
# Moles of MgO actually consumed = (1/2) x Moles of HCl consumed
= (1/2) x 0.100 mol
= 0.050 mol
Therefore, out of total 0.06958 mol of MgO taken, only 0.050 mol actually reacts with HCl. So, the total amount of energy released during the reaction is due to reaction of 0.050 mol MgO with HCl.
Step 3: Calculate molar enthalpy change:
Amount of energy released = - 9179.0512 J
Moles of MgO reacted = 0.050 mol
Now,
Molar enthalpy change for MgO = Total energy released / Moles of MgO consumed
= -9179.0512 J / 0.050 mol
= -183581.024 J/ mol
= -183.581 kJ/ mol
Hence, dHrxn / mol = -183.581 kJ/ mol
Note: Please recheck your calculated value of dH 25044.
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