Answer:
i) The number of moles of CO₂ (g) produced from the reaction = 0.07663 mole
ii) The volume of C₂H₅OH (l), in mL, that was combusted to produce the volume of CO₂ (g)
collected = 2.234 mL
iii) The amount of heat, in KJ, that was released by the combustion reaction = 52.4 kJ
Explanation:
The balanced chemical reaction when ethanol is combusted is given as
C₂H₅OH (l) + 3O₂ (g) → 2CO₂ (g) + 3H₂O (g)
The volume of CO₂(g) produced was 18.0 L when measured at 21.7°C and 1.03 atm.
i) Number of moles of CO₂ (g) produced by the reaction
With the correct and logical assumption that CO₂ is an ideal gas, the ideal gas equation has the relation
PV = nRT
P = pressure = 1.03 atm = 1.03 × 101325 Pa = 10,435.96 Pa
V = Volume of the gas = 18.0 L = 0.018 m³
n = number of moles = ?
R = molar gas constant = 8.314 J/mol.K
T = absolute temperature in Kelvin = 21.7 + 273.15 = 294.85 K
(10,435.96 × 0.018) = n × 8.314 × 294.85
n = 0.076629106 = 0.07663 mole
ii) The volume of C₂H₅OH (l), in mL, that was combusted to produce the volume of CO₂(g)
collected.
Recall the stoichiometric balance of the reaction
C₂H₅OH (l) + 3O₂ (g) → 2CO₂ (g) + 3H₂O (g)
2 moles of CO₂ is obtained from 1 mole of C₂H₅OH
0.07663 mole of CO₂ will be obtained from (0.07663×1/2) mole of C₂H₅OH; that is, 0.03831 mole of C₂H₅OH.
But we can convert this number of moles used up to mass of C₂H₅OH produced
Mass = (Number of moles) × (Molar Mass)
Molar mass of C₂H₅OH = 46.07 g/mol
Mass of C₂H₅OH combusted from the reaction
= 0.03831 × 46.07 = 1.765 g
But density of C₂H₅OH = 0.79 g/mL
Density = (Mass)/(Volume)
Volume = (Mass)/(Density) = (1.765/0.79)
= 2.234 mL
iii) The amount of heat, in KJ, that was released by the combustion reaction.
The heat of combustion of C₂H₅OH at the temperature of the reaction = -1367.6 kJ/mol. (From literature)
1 mole of C₂H₅OH combusts to give 1367.6 kJ of heat
0.03831 mole of C₂H₅OH will give (0.03831×1367.6) = 52.39 kJ = 52.4 kJ
Hope this Helps!!!!
