and volumes and (3 pts for Data) Be sure to include units! Recorded Data tempera
ID: 550437 • Letter: A
Question
and volumes and (3 pts for Data) Be sure to include units! Recorded Data temperatures, etc. c) Use EqS4 to calculate 1inloules a Data Part il1 T Trial 2d Trial Avera Concentration of HCt: Vol HCI 208mal temperohure chonge (AT 95 500l so o Concentration of NH:2.03 Vol NHs Calculate AH in kJ/mol (hint: you just Convert to kJ, and divide by moles of limiting reagent H in J Highest Temp 35.9 | 35.4°C 12.2°C-3.1. Calc ar| volume equation for the reaction in Part I11 (HCI + NH,) solution Use Eq #2 to calculate the theoretical Standard Molar Enthalpy of Formation (Arron) for the reaction in Part l (int use the 11 values from Appendix 3 in your text). Calculations (Complete after you have collected all data) B) -1 Calculations Part lll moles HC moles NH Limiting reagent is: Show work for full creditExplanation / Answer
Data Part III
1st Trial
2nd Trial
Average
Concentration of HCl (M)
2.08
Volume of HCl (mL)
49.5
50.0
-
Initial temperature (Ti) (ºC)
23.7
22.3
Concentration of NH3 (M)
2.03
Volume of NH3 (mL)
50.0
48.0
-
Highest temperature (Tf) (ºC)
35.9
35.4
-
Calculated T (ºC)
12.2
13.1
12.65
Calculated total volume (mL)
99.5
98.0
98.75
Calculated mass solution
99.5 [we assume the solutions to be aqueous solutions and the density of water is 1 g/mL; hence mass of solution = (total mass of water)*(density of water)]
98.0
98.75
Calculations Part III
1st. Trial
2nd. Trail
Average
Moles HCl
0.10296 (check calculation 1)
0.104
0.10348
Moles NH3
0.1015 (check calculation 2)
0.09744
0.09947
Limiting reagent is
NH3 (check calculation 3)
Calculation 1:
Moles HCl = (volume of HCl in L)*(concentration of HCl in mol/L) = (49.5 mL)*(1 L/1000 mL)*(2.08 M)*(1 mol/L/1 M) = 0.10296 mole.
Calculation 2:
Moles NH3 = (volume of NH3 in L)*(concentration of NH3 in mol/L) = (50.0 mL)*(1 L/1000 mL)*(2.03 M)*(1 mol/L/1 M) = 0.1015 mole.
Calculation 3:
Write down the balanced chemical equation for the reaction.
NH3(aq) + HCl(aq) ----------> NH4Cl(aq)
As per the stoichiometric equation,
1 mole NH3 = 1 mole HCl
We have 0.10348 mole HCl reacting with 0.09947 mole NH3. Clearly, NH3 is present in lower amount (1:1 molar stoichiometry of reaction) and hence, NH3 is the limiting reactant.
C + D) I need to know the equation #4. In particular, I do not have data to calculate the calorimetric constant.
E) The molecular equation is written as
NH3(aq) + HCl(aq) ----------> NH4Cl(aq)
HCl and NH4Cl are electrolytes in aqueous solution and ionize into ions as below.
NH3(aq) + H+(aq) + Cl-(aq) ---------> NH4+(aq) + Cl-(aq)
Cancel out the common ions from both sides and write
NH3(aq) + H+(aq) ---------> NH4+(aq)
Data Part III
1st Trial
2nd Trial
Average
Concentration of HCl (M)
2.08
Volume of HCl (mL)
49.5
50.0
-
Initial temperature (Ti) (ºC)
23.7
22.3
Concentration of NH3 (M)
2.03
Volume of NH3 (mL)
50.0
48.0
-
Highest temperature (Tf) (ºC)
35.9
35.4
-
Calculated T (ºC)
12.2
13.1
12.65
Calculated total volume (mL)
99.5
98.0
98.75
Calculated mass solution
99.5 [we assume the solutions to be aqueous solutions and the density of water is 1 g/mL; hence mass of solution = (total mass of water)*(density of water)]
98.0
98.75
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