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

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.