At Westonci.ca, we provide clear, reliable answers to all your questions. Join our vibrant community and get the solutions you need. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
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]
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.
