# Conditions Temperature (°C)

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 Conditions Temperature (°C) Acid used with calcium oxide 22 Resulting mixture from reaction of calcium oxide and acid 29 Acid used with calcium carbonate 22 Resulting mixture from reaction of calcium carbonate and acid 23

Treatment of results

1. Calculate the # of moles of:
i) calcium carbonate used

Molecular mass of CaC03 ii) calcium Oxide used
Mass=1g RMM of CaO=40+16 gmol-1
Mole=Mass/RMM
1g/56gmol-1
Moles of CaO= 0.01786mol
1*CA +1*c + 3*0
=1*40+1*12+3*16
=40+12+48
=100gmol-1
No of moles=1g/100=
Mole =given/RMM =0.01M

1. Calculate the enthalpy change of the reaction involving
i) calcium oxide

Enthalpy change of CaO=
Mass=50 g(since 1g=1cm3)),c=4.18Jg-1°C-1(s.h.c. Of water), ∆T
∆T=30.0-25.0 =5.0°c

Enthalpy Change=mc∆T =50*4.18*5.0= -1,045J

1. Calcium Carbonate mass= 50 g(since 1g=1cm3),c=4.18Jg-1°C(s.h.c),

2. ∆T=26.5-25.0=1.5°c

Enthalpy change= mc∆T
=50*4.18*1.5= -313.5J

Given that the heat capacity of the solution is 4.18 J g-1 °C-1 and that 1 cm3 of aqueous solution has a mass of 1 g

1. Using your answers to questions 1 and 2, determine the enthalpy change of reaction for
i) one mole of calcium oxide

Enthalpy change of CaO=-1,045J,Moles of CaO=0.01786mol
Enthalpy change per mole=Enthalpy change/mole
=- 1.045J/0.01786mol
=-58,510.6Jmol. -59KJmol

Calcium carbonate of CaCO= -313.5J Moles of CaCO= 0.01 mol

Enthalpy change per mole =Enthalpy change/mole= -313.5/0.01 mol= -31,350 Jmol-1 /-31kJmol-1

1. Draw an energy cycle for the enthalpy change of the reaction CaO(s) + CO2(g) → CaCO3(s)

using the TWO BALANCED chemical equations for the reactions you have conducted.