Answer:
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
