When 2.50 g of sucrose (molar mass = 342.30 g/mol) reacts with excess oxygen in a constant volume calorimeter, the temperature of the calorimeter increases by 9.17 $\circ$ C. The heat capacity of the calorimeter is 4.90 kJ/ $\circ$ C. Determine the heat of combustion per mole of sucrose.

Question

Quizplus · Accepted Answer

Firstly, we need to calculate the heat released in the reaction:
q = mcΔT = 4.90 kJ/F1F1F1S1F1F1F10C × 9.17F1F1F1S1F1F1F10C = 44.942 kJ
This heat is released by the combustion of 2.50 g of sucrose. To find the heat of combustion per mole of sucrose, we need to convert the mass of sucrose to moles:
moles of sucrose = 2.50 g / 342.30 g/mol = 0.007307 mol
Now we can calculate the heat of combustion per mole of sucrose:
ΔHcombustion = q / n = 44.942 kJ / 0.007307 mol = -6150 kJ/mol
The negative sign indicates that the reaction is exothermic (i.e. releases heat). Therefore, the correct answer is D: (F1F1F1S1F1F1F106,150 kJ/mol).

When 250 G of Sucrose (Molar Mass = 342 ∘\circ∘ C the Heat Capacity of the Calorimeter Is 4

When 250 G of Sucrose (Molar Mass = 342 $\circ$ C the Heat Capacity of the Calorimeter Is 4