Find out the calorific value of Glucose `C_(6)H_(12)O_(6)+6O_(2)rarr 6CO_(2)+6H_(2)O , Delta H=-2900` kJ/mole

Explanation

To find the calorific value of glucose, we will follow these steps:

### Step 1: Understand the Reaction
The combustion of glucose can be represented by the following reaction:
\[ \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \rightarrow 6\text{CO}_2 + 6\text{H}_2\text{O} \]
The enthalpy change (\( \Delta H \)) for this reaction is given as \(-2900 \, \text{kJ/mol}\). This means that when 1 mole of glucose is completely combusted, 2900 kJ of energy is released.

### Step 2: Calculate the Molar Mass of Glucose
...