Question

calculate the molar enthalpy of the combustion of butane if 1.22 g of the fuel increased the temperature of a tin can calorimeter (22.5 g of tin and 200 mL of water) by 8.30 celsius

Answer 1

The molar enthalpy of the combustion is the amount of energy released when 1 mole of the substance combusts in excess oxygen:

$∆H = -Q/n_{C_4H_{10}}$ .

The energy released can be calculated using the specific heat capacities of tin $c_{tin}$ ​ (0.218 J/g°C) and of water $c_w$ (4.186 J/g°C). Assuming the density of water equal 1 g/mL, the mass of water is 200 g. Therefore:

$Q = (c_{tin}m_{tin}+c_{w}m_w)∆T$

$Q = (0.218·22.5+4.186·200)·8.30 = 6.99·10^3$ J.

The number of the moles of butane is its mass divided by its molar mass (58.12 g/mol):

$n = m/M = 1.22/58.12 = 0.0210$ mol.

Finally, the molar enthalpy of combustion of butane is:

$∆H = -\frac{6.99·10^3}{0.0210} = 3.33·10^5$ J/mol, or 333 kJ/mol

Answer: the molar enthalpy of the combustion of butane is 333 kJ/mol.