Ask Question
1 May, 18:15

6.94)

The combustion of what volume of ethane (C2H6),

measured at 23.0°C and 752 mmHg, would be re-

quired to heat 855 g of water from 25.0°C to 98.0°C?

+1
Answers (1)
  1. 1 May, 20:59
    0
    3.69 L

    Explanation:

    If the total heat of the combustion will be used to heat water, then it must be equal to the sensitive heat of that amount of water. The sensitive heat can be calculated by:

    Q = m*c*ΔT

    Where Q is the heat, m is the mass, c is the specific heat (4.18 J/gºC for water), and ΔT is the temperature change (final - initial).

    Q = 855*4.18 * (98.0 - 25.0)

    Q = 260,894.7 J

    Q = 261 kJ

    The reaction of combustion is:

    C₂H₆ (l) + 7/2 O₂ (g) → 2CO₂ (g) + 3H₂O (l)

    The physical states are that because of the conditions that are (23.0ºC and 752 mmHg ≅ 1 atm).

    The enthalpies of formation of the substances are:

    ΔH°f (C₂H₆, l) = - 84.7 kJ/mol

    ΔH°f (O₂, g) = 0

    ΔH°f (CO₂, g) = - 393.5 kJ/mol

    ΔH°f (H₂O, l) = - 285.8 kJ/mol

    The enthalpy change of the reaction is

    ΔH°rxn = ∑nxΔH°f (products) - ∑nxΔH°f (reactants)

    ΔH°rxn = [3 * (-285.8) + 2 * (-393.5) ] - [1 * (-84.7) ]

    ΔH°rxn = - 1559.7 kJ/mol

    So, the number of moles of ethane to lose the amount of heat absorbed by water is the heat divided by the enthalpy of reaction:

    n = 261/1559.7

    n = 0.1673 moles

    The molar mass of ethane is:

    2*12 g/mol of C + 6 * 1 g/mol of H = 30 g/mol

    The mass of ethane is the molar mass multiplied by the number of moles

    m = 30*0.1673 = 5.019 g

    The density of ethane at atmospheric conditions is 1.36 g/L, so the volume is the mass divided by the density:

    V = 5.019/1.36

    V = 3.69 L
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “6.94) The combustion of what volume of ethane (C2H6), measured at 23.0°C and 752 mmHg, would be re- quired to heat 855 g of water from ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers