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Part A

For the reaction:

[tex]\[ \text{Ti(s) + 2 F}_2\text{(g) → TiF}_4\text{(s)} \][/tex]

Compute the theoretical yield of the product (in grams) for the following initial amounts of reactants:

[tex]\[ 7.0 \, \text{g Ti, 7.0 \, g F}_2 \][/tex]

Express your answer using two significant figures.

[tex]\[ m = 11 \, \text{g} \][/tex]

Note: The limiting reagent of this reaction is F[tex]\(_2\)[/tex] as it yields 11.4 g of TiF[tex]\(_4\)[/tex] compared to 18.1 g of TiF[tex]\(_4\)[/tex] using the mass of Ti to calculate yield.

Part B

For the reaction:

[tex]\[ \text{Ti(s) + 2 F}_2\text{(g) → TiF}_4\text{(s)} \][/tex]

Compute the theoretical yield of the product (in grams) for the following initial amounts of reactants:

[tex]\[ 2.3 \, \text{g Ti, 1.8 \, g F}_2 \][/tex]

Express your answer using two significant figures.

[tex]\[ m = 2.9 \, \text{g} \][/tex]

Note: The limiting reagent of this reaction is F[tex]\(_2\)[/tex] as it yields 2.93 g of TiF[tex]\(_4\)[/tex] compared to 5.95 g of TiF[tex]\(_4\)[/tex] using the mass of Ti to calculate yield.

Part C

For the reaction:

[tex]\[ \text{Ti(s) + 2 F}_2\text{(g) → TiF}_4\text{(s)} \][/tex]

Compute the theoretical yield of the product (in grams) for the following initial amounts of reactants:

[tex]\[ 0.227 \, \text{g Ti, 0.296 \, g F}_2 \][/tex]

Express your answer using two significant figures.

[tex]\[ m = \][/tex]

(Note: The answer for Part C was not provided in the original text. Please calculate the theoretical yield based on the limiting reagent.)

Sagot :

GinnyAnswer
Let's analyze each step to compute the theoretical yield of TiF₄ for the given reactants: 0.227 g of Ti and 0.296 g of F₂.

1. Molecular Masses:
- Molar mass of Ti (Titanium): 47.87 g/mol
- Molar mass of F₂ (Fluorine gas, where F₂ consists of 2 Fluorine atoms): [tex]\(2 \times 18.998\)[/tex] g/mol = 37.996 g/mol
- Molar mass of TiF₄ (Titanium tetrafluoride): 47.87 g/mol (for Ti) + [tex]\(4 \times 18.998\)[/tex] g/mol (for 4 Fluorine atoms) = 123.862 g/mol

2. Given Masses:
- Mass of Ti = 0.227 g
- Mass of F₂ = 0.296 g

3. Calculate Moles of Each Reactant:
- Moles of Ti: [tex]\(\frac{0.227 \text{ g}}{47.87 \text{ g/mol}} = 0.004742\)[/tex] moles
- Moles of F₂: [tex]\(\frac{0.296 \text{ g}}{37.996 \text{ g/mol}} = 0.007790\)[/tex] moles

4. Determine Limiting Reactant:
- According to the balanced equation [tex]\( \text{Ti} + 2 \text{F}_2 \rightarrow \text{TiF}_4 \)[/tex]:
- 1 mole of Ti reacts with 2 moles of F₂ to produce 1 mole of TiF₄.
- Moles of Ti needed to react with available F₂: [tex]\( \frac{0.007790 \text{ moles of F}_2}{2} = 0.003895 \text{ moles of Ti needed} \)[/tex]
- Since 0.004742 moles of Ti are available, Ti is in excess. So, F₂ is the limiting reactant.

5. Calculate Moles of TiF₄ Produced:
- Based on the limiting reactant (F₂), the moles of TiF₄ produced: [tex]\( 0.007790 \text{ moles of F}_2 \div 2 = 0.003895 \)[/tex] moles of TiF₄.

6. Calculate the Mass of TiF₄ Produced (Theoretical Yield):
- Mass of TiF₄: [tex]\( 0.003895 \text{ moles} \times 123.862 \text{ g/mol} \approx 0.482 \text{ g}\)[/tex].

Thus, for part C:
The theoretical yield of TiF₄ when using 0.227 g of Ti and 0.296 g of F₂ is 0.48 grams (rounded to two significant figures).
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