At `25^(@)C`, buring `0.2 "mole" H_(2)` with `0.1` mole `O_(2)` to produce `H_(2)O(l)` in a bomb calorimeter (constant volume) raises the temperature of the apperaturs `0.88^(@)C`. When `0.01mol` toulene is burned in this calorimeter, the temperature is raised by `0.615^(@)C`. Calculate `DeltaH^(Theta)` combustion of toluene. `Delta_(f)H^(Theta) H_(2)O(l) =- 286 kJ mol^(-1)`.

Explanation

To calculate the standard enthalpy change of combustion (ΔH°) for toluene, we can follow these steps:

### Step 1: Calculate the heat released during the combustion of hydrogen
We know that the combustion of hydrogen gas (H₂) with oxygen (O₂) produces water (H₂O) and releases energy. The reaction is:
\[ 2 H_2(g) + O_2(g) \rightarrow 2 H_2O(l) \]
Given that the standard enthalpy of formation of water (ΔH_f°) is -286 kJ/mol, we can find the heat released when burning 0.2 moles of H₂.

**Calculation:**
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