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Consider the reaction C12H22O11 (s) + 12 O2 (g) → 12 CO2 (g) + 11 H2O (l) in which 10.8 g of sucrose, C12H22O11, was burned in a bomb calorimeter with a heat capacity of 7.50 kJ/oC (including its water). The temperature inside the calorimeter was found to increase by 20.5 oC. Based on this information, what is the heat of this reaction per mole of sucrose? Enter your answer numerically, in terms of kJ/

User Nanu
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Answer:

THE HEAT OF THIS REACTION PER MOLE OF SUCROSE IS 4868.75 KJ OF HEAT.

Step-by-step explanation:

To answer this question:

First calculate the total heat given off by sucrose:

Total energy/ heat = heat capacity * change in temperature

Heat capacity = 7.50 kJ/ °C

Change in temperature = 20.5 °C

Heat = 7.50 kJ * 20.5 °C

Heat = 153.75 kJ of heat.

Next is to calculate the heat of reaction per mole of the sucrose

Equation of the reaction:

C12H22011 (s) + 12 O2 (g) ---------> 12 CO2 (g) + 11 H20(l)

Since 1 mole of sucrose will be the molar mass of sucrose, then we should calculate the molar mass of sucrose.

Molar mass of sucrose = ( 12* 12 + 1 * 22+ 16*11) g/mol

Molar mass = 342 g/mol of sucrose

Since 10.8 g of sucrose produces 153.75 kJ of heat, 342 g will produces how many joules of heat?

10.8 g of sucrose = 153.75 kJ of heat

342 g of sucrose = ( 342 * 153.75 kJ / 10.8)

= 52 582.5 kJ / 10.8

= 4868.75 kJ of heat

So therefore, 1 mole of sucrose will produce 4868.75 kJ of heat.