<u>Answer:</u> The amount of heat released is -7.203 kJ
<u>Explanation:</u>
The given chemical equation follows:
![2H_2O_2(l)\rightarrow 2H_2O(l )+O_2(g);\Delta H=-196kJ](https://tex.z-dn.net/?f=2H_2O_2%28l%29%5Crightarrow%202H_2O%28l%20%29%2BO_2%28g%29%3B%5CDelta%20H%3D-196kJ)
To calculate the enthalpy change for 1 mole of the hydrogen peroxide, we use unitary method:
When 2 moles of hydrogen peroxide is reacted, the enthalpy of the reaction is -196 kJ
So, when 1 mole of hydrogen peroxide will react, the enthalpy of the reaction will be ![\frac{-196}{2}\times 1=-98kJ](https://tex.z-dn.net/?f=%5Cfrac%7B-196%7D%7B2%7D%5Ctimes%201%3D-98kJ)
- To calculate the number of moles, we use the equation:
![\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%7D%7D%7B%5Ctext%7BMolar%20mass%7D%7D)
Given mass of hydrogen peroxide = 2.50 g
Molar mass of hydrogen peroxide = 34 g/mol
Putting values in above equation, we get:
![\text{Moles of hydrogen peroxide}=\frac{2.50g}{34g/mol}=0.0735mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20hydrogen%20peroxide%7D%3D%5Cfrac%7B2.50g%7D%7B34g%2Fmol%7D%3D0.0735mol)
- To calculate the heat of the reaction, we use the equation:
![\Delta H_{rxn}=\frac{q}{n}](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Cfrac%7Bq%7D%7Bn%7D)
where,
= amount of heat released
n = number of moles = 0.0735 moles
= enthalpy change of the reaction = -98 kJ/mol
Putting values in above equation, we get:
![-98kJ/mol=\frac{q}{0.0735mol}\\\\q=(-98kJ/mol\times 0.0735mol)=-7.203kJ](https://tex.z-dn.net/?f=-98kJ%2Fmol%3D%5Cfrac%7Bq%7D%7B0.0735mol%7D%5C%5C%5C%5Cq%3D%28-98kJ%2Fmol%5Ctimes%200.0735mol%29%3D-7.203kJ)
Hence, the amount of heat released is -7.203 kJ