The combustion of titanium with oxygen produces titanium dioxide: ti(s) + o2(g)

The heat of reaction for the combustion of a mole of Ti in this calorimeter is - 1.52 × 10^4 KJ/mol.

The equation of the reaction is;

Ti(s) + O2(g) → TiO2(s)

From the available information;

Temperature difference(θ) =  91.60 °C -  25.00 °C = 66.6°C

heat capacity of calorimeter (Cp) = 9.84 kJ/K

Heat absorbed by the calorimeter = cpθ = 9.84 × 10^3 J/K × 66.6°C

= 655.3 KJ

Number of moles of Ti = 2.0600 / 47.867 = 0.0430 mol

Enthalpy of combustion in KJ/mol = -(655.3 KJ)/0.0430 mol

= - 1.52 × 10^4 KJ/mol

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64800 kJ/mol

Explanation:

Heat gained by bomb calorimeter =Q

Heat capacity of bomb calorimeter , C =9.84 kJ/K =

Change in temperature = ΔT= (91.60-25.00) °C = 66.6°C  = 66.6 K

Let the heat released during reaction be q.

Total heat released during reaction is equal to total heat gained by water and bomb calorimeter.

Thus 2.060 g of titanium releases = 655.34 kJ of heat

1 g of titanium releases = of heat

1 mole or 48 g of titanium releases = of heat

Thus heat given off by the burning titanium, in kJ/mol is 64800.

d i think hope that helps