Thermite mixtures are used for certain types of weldings and the thermite reaction is highly exothermic. Given delta H= -852 KJ/mol, 1 mole of granular Fe2O3 and 2 moles of granular aluminum are mixed at room temperature and the reaction is initiated. the liberated heat is retain within the products whose combined specific heat over a broad temperature range is about 0.8 J/g C . the melting pt. of Fe is 1530 C. show that the quantity heat liberated is more than sufficient to raise the temp. of the products to the melting pt. of Fe.

Fe2O3 + 2Al - Al203 + 2Fe

To calculate whether the quantity of heat liberated is sufficient to raise the temperature of the products to the melting point of Fe (1530°C), we need to determine the amount of heat produced in the thermite reaction.

1. Calculate the moles of Fe2O3 and Al:
- 1 mole of Fe2O3
- 2 moles of Al

2. Calculate the heat released in the reaction:
- ΔH = -852 KJ/mol

3. Convert the moles of Fe2O3 to grams:
- Molar mass of Fe2O3 = (2 × atomic mass of Fe) + (3 × atomic mass of O)
- Atomic mass of Fe = 55.85 g/mol
- Atomic mass of O = 16.00 g/mol
- Molar mass of Fe2O3 = (2 × 55.85) + (3 × 16.00) = 159.69 g/mol
- Mass of Fe2O3 = 159.69 g/mol

4. Convert the moles of Al to grams:
- Molar mass of Al = 26.98 g/mol
- Mass of Al = (26.98 g/mol) × 2 moles = 53.96 g

5. Calculate the total mass of the reactants:
- Total mass = mass of Fe2O3 + mass of Al = 159.69 g + 53.96 g = 213.65 g

6. Calculate the heat released per gram of reactant:
- ΔH per gram = ΔH / total mass
- ΔH per gram = -852 KJ/mol / 213.65 g = -3.989 KJ/g

7. Determine the temperature change:
- Specific heat (c) of the products = 0.8 J/g°C
- Initial temperature = room temperature (assume 25°C)
- Final temperature = melting point of Fe (1530°C)
- Temperature change = final temperature - initial temperature = 1530°C - 25°C = 1505°C

8. Calculate the heat required to raise the temperature to the melting point:
- Heat required = mass of products × specific heat × temperature change
- Mass of products = mass of Fe2O3 + mass of Al = 213.65 g
- Heat required = 213.65 g × 0.8 J/g°C × 1505°C = 257,164 J

The quantity of heat liberated, which can be calculated using the molar enthalpy, is more than sufficient to raise the temperature of the products to the melting point of Fe.