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.

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The state or quality of being humid.a quantity representing the amount of water vapor in the atmosphere or a gas.