Question

Calculate the heat associated with the consumption of 1.542 mol of O2 in this reaction.

Calculate the heat associated with combustion of 25.08 g of butane.

Calculate the mass of butane that must be burned in order to heat 23.83 kg of water from 22.31 ∘C to 65.71 ∘C. Assume no loss of heat in the transfer from the reaction to the water. The specific heat of water is 4.184 J/g∘C.

Consider the following thermochemical equation for the combustion of butane.
2C4H10(g)+15O2(g)→8CO2(g)+10H2O(g)ΔH∘rxn=−5314.6kJ

Answer 1

2C4H10(g)+15O2(g) → 8CO2(g)+10H2O(g) ΔH∘rxn = −5314.6kJ

a) 15 mol O2 = -5314.6 kj

so that,

1.542 mol O2 on consumption = -5314*1.542/15 = -546.3 kj

b)

No of mol of butane = 25.08 / 58 = 0.432 mol

so that

amount of energy released = 0.432*5314/2 = 1147.824 kj

c)

heat absorbed by water = m*s*DT

    = 23.83*10^3 *4.18*(65.71-22.31)

    = 4323.05 kj

2 mol butane = - 5314.6 kj

No of mol of butane = 4323.05*2/5314 = 1.63 mol

mass of butane = 1.63*58 = 94.54 grams