When a 5.000 g mixture of CaCO3, Ca(HCO3)2, CaCl2 and Ca(ClO3)2 is heated at elevated temperature ga — Physical Chemistry — Thermodynamics Chemistry Question
Analyzing a mixture of calcium salts
When a 5.000 g mixture of CaCO3, Ca(HCO3)2, CaCl2 and Ca(ClO3)2 is heated at elevated temperature gaseous CO2, H2O, and O2 are evolved. The gases evolved exert a pressure of 1.312 atm in an evacuated 1.000 dm–3 cylinder at 400.0 K. When the temperature inside the cylinder is decreased to 300.0 K, the pressure drops to 0.897 atm. The vapor pressure of water at this temperature is 27.0 torr. The gas in the cylinder is used to combust an unknown amount of acetylene C2H2. The enthalpy change during the combustion process is determined as –7.796 kJ with the use of a calorimeter.
∆fH°(C 2H2(g)) = 226.8 kJ mol–1; ∆fH°(CO 2(g)) = –393.5 kJ mol–1;
∆fH°(H 2O(g)) = –241.8 kJ mol–1; ∆vapH°298K(H2O(l)) = 44.0 kJ mol–1
Write balanced equations for the possible decomposition reactions generating gases.
Model Answer
CaCO3(s) → CaO(s) + CO2(g)
Ca(HCO3)2(s ) → CaO(s) + 2 CO2(g) + H2O(g)
CaCl2 → no reaction
Ca(ClO3)2(s) → CaCl2(s) + 3 O2(g)
Write a balanced equation for the combustion of C2H2.
Model Answer
2 C3H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(l)
Calculate the amounts of substances (in moles) of gases produced in the cylinder.
Model Answer
ntotal = p V / (R T) = 1.31 × 1.000 / (0.082 × 400) = 0.0399 mol = 0.040 mol
Calculate the amount of substance of O2 that was present in the cylinder.
Model Answer
Δr H° = 2 Δf H°(CO2(g)) + Δf H°(H2O(l)) - Δf H°(C2H2(g))
H2O(g) → H2O(l)
Δvap H° = Δf H°(H2O(l)) - Δf H°(H2O(g))
Δf H°(H2O(l)) = -44.0 + (-241.8) = -285.8 kJ mol-1
n(O2) = 5/2 n(C2H2) = 5/2 × (– 7.796) / (– 1299.6) = 0.015 mol
Calculate the amounts of substances of CO2 and H2O produced.
Model Answer
At 300 K, H2O(g) condenses.
p(CO2 + O2) = ptotal – p(H2O) = 0.897 – 27.0 / 760 = 0.861 atm
n(CO2 + O2) = p V / (R T) = 0.861 × 1.00 / (0.082 × 300) = 0.035 mol
n(CO2) = 0.035 – 0.015 = 0.020 mol
n(H2O) = 0.040 – 0.035 = 0.005 mol
Calculate the mass percentage of CaCO3 and CaCl2 in the original mixture.
Model Answer
n(Ca(HCO3)2) = n(H2O) = 0.005 mol
n(Ca(ClO3)2) = 1/3 n(O2) = 1/3 × 0.015 = 0.005 mol
n(CaCO3) = n(CO2)CaCO3
n(CO2) = n(CO2)CaCO3 + n(CO2)Ca(HCO3)2
n(CO2) = n(CO2)CaCO3 + 2 × n(H2O)
0.0200 = n(CO2)CaCO3 + 2 × 0.0050
n(CaCO3) = n(CO2)CaCO3 = 0.020 – 0.010 = 0.010 mol
m(Ca(HCO3)2) = 0.0050 × 162.110 = 0.81 g
m(Ca(ClO3)2) = 0.0050 × 206.973 = 1.03 g
m(CaCO3) = 0.0100 × 100.086 = 1.001 g
m(CaCl2) = 5.000 – (0.8106 + 1.034 + 1.001) = 2.153 g
% CaCl2 = 2.153 / 5.000 × 100 = 43.0 %
% CaCO3 = 1.001 / 5.000 × 100 = 20.0 %