Question

Using a molar heat of combustion of hexane (C6H14) –4163 kJ/mol, the mass of hexane required to increase the temperature of 1.76 L of water from 21.0°C to 70.0°C is a.bc g.

Put your answer for a.bc in the blank

Answer 1

Answer:

7.46 g.

Explanation:

• Firstly, we need to calculate the amount of heat needed to warm 5.64 kg of water from 21.0°C to 70.0°C using the relation:

Q = m.c.ΔT,

where, Q is the amount of heat absorbed by water (Q = ??? J).

m is the mass of water (m: we will determine).

c is the specific heat capacity of water (c = 4.186 J/g.°C).

ΔT is the temperature difference (final T - initial T) (ΔT =  70.0 °C - 21.0 °C = 49.0 °C).

• To determine the mass of 1.76 L of water we can use the relation:

mass = density x volume.

density of water = 1000 g/L & V = 1.76 L.

∴ mass = density x volume = (1000 g/L)(1.76 L) = 1760.0 g.

∵ Q = m.c.ΔT

∴ Q = m.c.ΔT = (1760.0 g)(4.186 J/g.°C)(49.0 °C) = 360483.2 J ≅ 360.4832 kJ.

• As mentioned in the problem the molar heat of combustion of hexane is - 4163.0 kJ/mol.

Using cross multiplication we can get the no. of moles of hexane that are needed to be burned to release 360.4832 kJ:

Combustion of 1.0 mole of methane releases → - 4163.0 kJ.

Combustion of ??? mole of methane releases → - 360.4832 kJ.

∴ The no. of moles of hexane that are needed to be burned to release 360.4832 kJ = (- 360.4832 kJ)(1.0 mol)/(- 4163.0 kJ) = 0.0866 mol.

• Now, we can get the mass of hexane that must be burned to warm 1.76 L of water from 21.0°C to 70.0°C:

∴ mass = (no. of moles needed)(molar mass of hexane) = (0.0866 mol)(86.18  g/mol) = 7.46 g.