A35.6 g sample of ethanol (c2h5oh) is burned in a bomb calorimeter, according to the following reaction. if the temperature rose from 35.0 to 76.0°c and the heat capacity of the calorimeter is
23.3 kj/°c, what is the value of dh°rxn? the molar mass of ethanol is 46.07 g/mol.

c2h5oh(l) + o2(g) → co2(g) + h2o(g) δh°rxn = ? (points : 1)
(a) -1.24 × 103 kj/mol
(b) +1.24 × 103 kj/mol
(c) -8.09 × 103 kj/mol
(d) -9.55 × 103 kj/mol
(e) +9.55 × 103 kj/mol

A) -1.24 × 10^3 kJ/Mol

we are given;

Mass of ethanol, m = 35.6 g

Temperature change, Δt(35.0 to 76.0°C) = 41 °C

Specific heat capacity of the calorimeter, c = 23.3 kJ/°C

Molar mass of ethanol = 46.07 g/mol

We are required to the heat change of the reaction.

Therefore; we are going to use the following steps;

We know, Moles = Mass ÷ molar mass

Thus, moles of ethanol = 35.6 g ÷ 46.07 g/mol

                                      = 0.773 moles

Heat change = -mcΔt

but we are given s[pecific heat capacity in Kj/°C and we require heat change in kJ/mol

Therefore;

Heat change = -(cΔt) ÷ n ( n is the number of moles)

                      = -( 23.3 kJ/°C × 41°C) ÷0.773 mol

                    = - 1.24 × 10^3 kJ/Mol

Therefore, values of ΔH of the reaction is -1.24 × 10^3 kJ/Mol

can you attach the