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(b) When a sample of C2H5OH was combusted, the volume of CO2(g) produced was 18.0 L when measured at
21.7°C and 1.03 atm.
(i) Determine the number of moles of CO2(g) that was produced.
(ii) Determine the volume of C2H5OH(1), in mL, that was combusted to produce the volume of CO2(g)
collected. (The density of C2H5OH() is 0.79 g/mL.)
(iii) Determine the amount of heat, in KJ, that was released by the combustion reaction.

Respuesta :

Answer:

i) The number of moles of COâ‚‚ (g) produced from the reaction = 0.07663 mole

ii) The volume of Câ‚‚Hâ‚…OH (l), in mL, that was combusted to produce the volume of COâ‚‚ (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 COâ‚‚ (g) produced by the reaction

With the correct and logical assumption that COâ‚‚ 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 Câ‚‚Hâ‚…OH (l), in mL, that was combusted to produce the volume of COâ‚‚(g)

collected.

Recall the stoichiometric balance of the reaction

C₂H₅OH (l) + 3O₂ (g) → 2CO₂ (g) + 3H₂O (g)

2 moles of COâ‚‚ is obtained from 1 mole of Câ‚‚Hâ‚…OH

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 Câ‚‚Hâ‚…OH produced

Mass = (Number of moles) × (Molar Mass)

Molar mass of Câ‚‚Hâ‚…OH = 46.07 g/mol

Mass of Câ‚‚Hâ‚…OH combusted from the reaction

= 0.03831 × 46.07 = 1.765 g

But density of Câ‚‚Hâ‚…OH = 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 Câ‚‚Hâ‚…OH at the temperature of the reaction = -1367.6 kJ/mol. (From literature)

1 mole of Câ‚‚Hâ‚…OH 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!!!!

pstnonsonjoku

The heat evolved by the combustion of ethanol according to the question is -2105.2 kJ.

The equation of the reaction is;

C₂H₅OH (l) + 3O₂ (g) → 2CO₂ (g) + 3H₂O (g)

i) Using;

PV = nRT

P = 1.03 atm

n = ?

R = 0.082 atmLK-1mol-1

V = 18.0 L

T = 21.7°C + 273=  294.7 K

n = PV/RT

n =  1.03 atm × 18.0 L/0.082 atmLK-1mol-1 × 294.7 K

n = 18.54/24.17

n = 0.77 moles

ii) 1 mole of ethanol yields 2 moles of carbon dioxide

0.77 moles of ethanol yields 0.77 moles × 2 moles/1 mole

= `1.54 moles

Molar mass of ethanol = 46 g/mol

Mass of ethanol = `1.54 moles × 46 g/mol = 70.84 g

Volume of ethanol = mass/ density = 70.84 g/0.79 g/mL = 89.7 mL

iii) The combustion of 1 mole of ethanol releases  -1367 kJ

Combustion of `1.54 moles of ethanol releases `1.54 moles ×  -1367 kJ/ 1 mole

= -2105.2 kJ

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