To calculate the atom economy for each reaction, we need to utilize the formula for atom economy:
[tex]\[ \text{Atom Economy} = \left( \frac{\text{Molar Mass of Desired Product}}{\text{Total Molar Mass of Reactants}} \right) \times 100 \][/tex]
### Reaction 1: [tex]\( TiO_2 + 2 Mg \rightarrow Ti + 2 MgO \)[/tex]
1. Identify the desired product: In this reaction, the desired product is Titanium (Ti).
2. Calculate the molar mass of the desired product:
- Molar mass of Ti = 47.87 g/mol.
3. Calculate the total molar mass of the reactants:
- Molar mass of [tex]\( TiO_2 \)[/tex] = 47.87 (Ti) + 2 \times 16 (O) = 47.87 + 32 = 79.87 g/mol.
- Molar mass of 2 Mg = 2 \times 24.305 = 48.61 g/mol.
- Total molar mass of reactants = 79.87 (molar mass of [tex]\( TiO_2 \)[/tex]) + 48.61 (molar mass of 2 Mg) = 128.48 g/mol.
4. Calculate the atom economy:
[tex]\[ \text{Atom Economy} = \left( \frac{47.87}{128.48} \right) \times 100 = 37.26\% \][/tex]
### Reaction 2: [tex]\( TiO_2 \rightarrow Ti + O_2 \)[/tex]
1. Identify the desired product: In this reaction, the desired product is also Titanium (Ti).
2. Calculate the molar mass of the desired product:
- Molar mass of Ti = 47.87 g/mol.
3. Calculate the total molar mass of the reactants:
- Molar mass of [tex]\( TiO_2 \)[/tex] = 47.87 (Ti) + 2 \times 16 (O) = 47.87 + 32 = 79.87 g/mol.
4. Calculate the atom economy:
[tex]\[ \text{Atom Economy} = \left( \frac{47.87}{79.87} \right) \times 100 = 59.93\% \][/tex]
### Conclusion:
The atom economies for the two reactions are:
- For Reaction 1: 37.26%
- For Reaction 2: 59.93%
Based on the percentage options given for the displacement method:
The approximate atom economy for the displacement method ([tex]\( TiO_2 + 2 Mg \rightarrow Ti + 2 MgO \)[/tex]) is closest to Option D) 25%.
