Please explain how u got the answers. Thank you in advance! 1) A) A 65.45-g piec
ID: 878586 • Letter: P
Question
Please explain how u got the answers. Thank you in advance!
1)
A) A 65.45-g piece of zinc metal at 75.0°C was dropped into a calorimeter containing 125 mL water at 19.9 °C. If the apparatus comes to thermal equilibrium at 22.2 °C, what is the heat capacity of the calorimeter?
In the same calorimeter, 5.00-g of NH4Cl was dissolved in 150.0 g of water at 20.7 °C. The temperature of the solution dropped to 18.9 °C as the ammonium chloride finished dissolving. B) What is the heat consumed by this reaction, q (J)?
C) What is the heat consumed by one mole dissolving, delta H (kJ/mol)?
2) A 0.712-g sample of magnesium is burned in excess oxygen in a bomb calorimeter with a heat capacity of 722 J/C° containing 350.0 g of water initially at 21.57°C. The temperature of the apparatus rises to 26.32°C.
A) What is the enthalpy of combustion (in kJ/mol) of magnesium?
B)What is the heat of reaction in kJ for the balanced equation: 2 Mg(s) + O2(g) -------> 2 MgO(s)
3) Consider the reaction: 2HCl(aq) + Ba(OH)2(aq) ----------> BaCl2(aq) + 2 H2O(l) delta H = -118 kJ
A) Calculate the heat produced when 700.0 mL of 0.500 M HCl is mixed with 300.0 mL of 0.500 M Ba(OH)2. Note: One of the solutions is the limiting reagent.
B) Calculate the final temperature if both solutions were initially at 25.0° C, and they were mixed together in a calorimeter with a heat capacity of 180.3 J/C° Note: The density specific heat capacity of the solution are the same as that of water: D = 1.00 g/cm3 C = 4.184 J /g•°C
Explanation / Answer
From literature
Specific heat of Zinc =0.376 J/g.Deg.C
Specific heat of water = 4.187 J/g.Deg.C
heat given up by Zinc metal =Mass * Specific heat * Temperature difference = 65,45*0.376*(75-22.2)=1304.288 J
Water density is assuemd as 1g/cc
Heat taken by water = 125ml *(1g/cc) *4.187* (22.2-19.9)=1201.75 J
assuming adiabatic system, the baalnce heat has to be taken by Calorimeter
Heat taken by Calorimeter = 1304.288-1201.75=102.5376 J= (22.2-19.9)* heat capacity
Heat capacity= 102.5376/(22.2-19.9) =44.5817 J/deg.c
b) Heat given by water = 150*4.18*(20.7-18.9) = 1128.6 J
Heat taken by Calorimeter =44.5817*(20.7-18.9) = 80.24 Joules
Heat taken by Reaction = 1128.6- 80.24=1048.36 Joules
Molecular weight of NH4Cl =14+4+35.5 =53.5
Moles of NH4Cl = 5/53.5 =0.093 moles
0.093 moles give rise to 1048.36 joues
0.093 gives rise to 1.043836 Kj
1 moles 1.04836/ 0.093=11.27 Kj/mol
2.) Heat taken by Water =-350*4.18*(26.32-21.57)=-6949.2 Joules
Heat taken by Calorimeter=- 722*(26.32- 21.57) = -3429.5 Joules
Heat generated=- 6949.2 - 3429.5=-10378.5 Joules. This much heat is given by heat of combustion
Moles of Magnesium = 0.712/24 ( 24 is atomic weight of Mg) = 0.029667 moles
Heat of combustion= -10378.5/ 0.029667= 349845. Joules/ gmoles= -349.845 Kj/Mol
This much heat has to be produced by burning of Mg by
2Mg+O2---> 2MgO
Hence, Heat of formation
Heat of formation of products- Heat of formation of reactants
Since heat of formation of reactants is Zero
Heat of formation = 2* -349.845=-699 Kj
3. The reaction
2HCl+ Ba(OH)2 ---> BaCl2+ 2H2O DelH= -118Kj
Moles of HCl given =700ml of 0,.5M = (700/1000)*0.5=0.35 Moles
Moles of Ba(OH)2 = (300/1000)* 0.5 =0.15 moles
Stoichiometry suggests Ratio of HCl : Ba(OH)2 : 2:1
Actutal ratio : 0,35 : 0.15 =2.33 :1
Hence, HCl is excess reactant. Ba(OH)2 is the Limiting reactant permits the amount of BaCl2
Moles of Ba(OH)2 = 0.15 moles
DelH for reaction = -118 Kj when 1 moles of Ba(OH)2 is reacted
0.15 moles give rise to 0.15*-118 Kj=17.7 Kj =-17700 Joules
The final temperature can be calculated by balance= Heat taken by Calorimeter + heat taken by HCl and Ba(OH)2
17700= 180.3 *( T-25) + 700*4.18*(T-25)+ 300*1*4.18*(T-25)= (180.3+4180)*(T-25)
T-25 =17700/ 4360.3= 4.059
T= 25+4.059= 29.059 Deg.C