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Sagot :
To determine the theoretical yield for the reaction
[tex]\[ \text{Ti (s)} + 2\text{F}_2\text{(g)} \rightarrow \text{TiF}_4\text{(s)} \][/tex]
with initial amounts of 7.0 grams of titanium (Ti) and 7.0 grams of fluorine gas (F[tex]\(_2\)[/tex]), follow these steps:
### Step 1: Determine the Moles of Each Reactant
Moles of Ti:
1. Find the molar mass of Ti, which is 47.867 g/mol.
2. Use the given mass of Ti, which is 7.0 grams.
3. Calculate the moles of Ti:
[tex]\[ \text{moles of Ti} = \frac{7.0 \text{ g}}{47.867 \text{ g/mol}} \approx 0.146 \text{ moles} \][/tex]
Moles of F[tex]\(_2\)[/tex]:
1. Find the molar mass of F[tex]\(_2\)[/tex]. Note that each fluorine atom has a molar mass of 18.9984 g/mol, so:
[tex]\[ \text{molar mass of F}_2 = 2 \times 18.9984 \text{ g/mol} = 37.9968 \text{ g/mol} \][/tex]
2. Use the given mass of F[tex]\(_2\)[/tex], which is 7.0 grams.
3. Calculate the moles of F[tex]\(_2\)[/tex]:
[tex]\[ \text{moles of F}_2 = \frac{7.0 \text{ g}}{37.9968 \text{ g/mol}} \approx 0.184 \text{ moles} \][/tex]
### Step 2: Determine the Limiting Reactant
The balanced chemical equation shows that 1 mole of Ti reacts with 2 moles of F[tex]\(_2\)[/tex]:
- For Ti:
[tex]\[ \text{Moles of Ti} = 0.146 \text{ moles} \][/tex]
- For F[tex]\(_2\)[/tex]:
[tex]\[ \text{Moles of Ti that can react} = \frac{\text{moles of F}_2}{2} = \frac{0.184 \text{ moles}}{2} = 0.092 \text{ moles} \][/tex]
Since 0.092 moles (derived from F[tex]\(_2\)[/tex]) < 0.146 moles (derived from Ti), F[tex]\(_2\)[/tex] is the limiting reactant.
### Step 3: Calculate the Theoretical Yield of TiF[tex]\(_4\)[/tex]
The theoretical yield is based on the limiting reactant, F[tex]\(_2\)[/tex]:
1. Each mole of F[tex]\(_2\)[/tex] produces [tex]\( \frac{1}{2} \)[/tex] mole of TiF[tex]\(_4\)[/tex].
2. Therefore, the moles of TiF[tex]\(_4\)[/tex] produced:
[tex]\[ \text{Moles of TiF}_4 = 0.092 \text{ moles} \][/tex]
3. Find the molar mass of TiF[tex]\(_4\)[/tex]:
[tex]\[ \text{molar mass of TiF}_4 = 123.886 \text{ g/mol} \][/tex]
4. Calculate the mass of TiF[tex]\(_4\)[/tex]:
[tex]\[ \text{Mass of TiF}_4 = \text{moles of TiF}_4 \times \text{molar mass of TiF}_4 = 0.092 \text{ moles} \times 123.886 \text{ g/mol} \approx 11.4 \text{ grams} \][/tex]
### Final Answer:
The theoretical yield of TiF[tex]\(_4\)[/tex] is 11 grams, expressed to two significant figures. Therefore:
[tex]\[ m = 11 \text{ grams} \][/tex]
[tex]\[ \text{Ti (s)} + 2\text{F}_2\text{(g)} \rightarrow \text{TiF}_4\text{(s)} \][/tex]
with initial amounts of 7.0 grams of titanium (Ti) and 7.0 grams of fluorine gas (F[tex]\(_2\)[/tex]), follow these steps:
### Step 1: Determine the Moles of Each Reactant
Moles of Ti:
1. Find the molar mass of Ti, which is 47.867 g/mol.
2. Use the given mass of Ti, which is 7.0 grams.
3. Calculate the moles of Ti:
[tex]\[ \text{moles of Ti} = \frac{7.0 \text{ g}}{47.867 \text{ g/mol}} \approx 0.146 \text{ moles} \][/tex]
Moles of F[tex]\(_2\)[/tex]:
1. Find the molar mass of F[tex]\(_2\)[/tex]. Note that each fluorine atom has a molar mass of 18.9984 g/mol, so:
[tex]\[ \text{molar mass of F}_2 = 2 \times 18.9984 \text{ g/mol} = 37.9968 \text{ g/mol} \][/tex]
2. Use the given mass of F[tex]\(_2\)[/tex], which is 7.0 grams.
3. Calculate the moles of F[tex]\(_2\)[/tex]:
[tex]\[ \text{moles of F}_2 = \frac{7.0 \text{ g}}{37.9968 \text{ g/mol}} \approx 0.184 \text{ moles} \][/tex]
### Step 2: Determine the Limiting Reactant
The balanced chemical equation shows that 1 mole of Ti reacts with 2 moles of F[tex]\(_2\)[/tex]:
- For Ti:
[tex]\[ \text{Moles of Ti} = 0.146 \text{ moles} \][/tex]
- For F[tex]\(_2\)[/tex]:
[tex]\[ \text{Moles of Ti that can react} = \frac{\text{moles of F}_2}{2} = \frac{0.184 \text{ moles}}{2} = 0.092 \text{ moles} \][/tex]
Since 0.092 moles (derived from F[tex]\(_2\)[/tex]) < 0.146 moles (derived from Ti), F[tex]\(_2\)[/tex] is the limiting reactant.
### Step 3: Calculate the Theoretical Yield of TiF[tex]\(_4\)[/tex]
The theoretical yield is based on the limiting reactant, F[tex]\(_2\)[/tex]:
1. Each mole of F[tex]\(_2\)[/tex] produces [tex]\( \frac{1}{2} \)[/tex] mole of TiF[tex]\(_4\)[/tex].
2. Therefore, the moles of TiF[tex]\(_4\)[/tex] produced:
[tex]\[ \text{Moles of TiF}_4 = 0.092 \text{ moles} \][/tex]
3. Find the molar mass of TiF[tex]\(_4\)[/tex]:
[tex]\[ \text{molar mass of TiF}_4 = 123.886 \text{ g/mol} \][/tex]
4. Calculate the mass of TiF[tex]\(_4\)[/tex]:
[tex]\[ \text{Mass of TiF}_4 = \text{moles of TiF}_4 \times \text{molar mass of TiF}_4 = 0.092 \text{ moles} \times 123.886 \text{ g/mol} \approx 11.4 \text{ grams} \][/tex]
### Final Answer:
The theoretical yield of TiF[tex]\(_4\)[/tex] is 11 grams, expressed to two significant figures. Therefore:
[tex]\[ m = 11 \text{ grams} \][/tex]
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