Titanium , which is used to make air plane engines and frames, can be obtained from titanium tetrachloride , which in turn is obtained from titanium oxide by the following process :
`3 TiO_(2)(s) + 4C(s) + 6Cl_(2)(g) to 3TiCl_(4)(g) + 2CO_(2)(g) + 2CO (g)`
A vessel contains 4.3 g `TiO_(2),5.76 g C and ,7.1 g Cl_(2)` ,suppose the reaction goes to completion as written , how many gram of `TiCl_(4)` can be produced ? (Ti =48)

Answer

To solve the problem of how many grams of titanium tetrachloride (TiCl₄) can be produced from the given amounts of titanium dioxide (TiO₂), carbon (C), and chlorine (Cl₂), we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ 3 \text{TiO}_2(s) + 4 \text{C}(s) + 6 \text{Cl}_2(g) \rightarrow 3 \text{TiCl}_4(g) + 2 \text{CO}_2(g) + 2 \text{CO}(g) \]

### Step 2: Calculate the molar masses of the reactants
- Molar mass of TiO₂:
- Titanium (Ti) = 48 g/mol
- Oxygen (O) = 16 g/mol
- Molar mass of TiO₂ = 48 + (2 × 16) = 48 + 32 = 80 g/mol

- Molar mass of C (Carbon) = 12 g/mol

- Molar mass of Cl₂ (Chlorine) = 2 × 35.5 = 71 g/mol

### Step 3: Calculate the number of moles of each reactant
- Moles of TiO₂:
\[ \text{Moles of TiO}_2 = \frac{4.3 \text{ g}}{80 \text{ g/mol}} = 0.05375 \text{ moles} \]

- Moles of C:
\[ \text{Moles of C} = \frac{5.76 \text{ g}}{12 \text{ g/mol}} = 0.48 \text{ moles} \]

- Moles of Cl₂:
\[ \text{Moles of Cl}_2 = \frac{7.1 \text{ g}}{71 \text{ g/mol}} = 0.1 \text{ moles} \]

### Step 4: Determine the stoichiometric ratios
From the balanced equation:
- 3 moles of TiO₂ react with 4 moles of C and 6 moles of Cl₂.

Now, we will calculate the ratio of moles of each reactant to its stoichiometric coefficient:
- For TiO₂:
\[ \frac{0.05375}{3} = 0.01792 \]
- For C:
\[ \frac{0.48}{4} = 0.12 \]
- For Cl₂:
\[ \frac{0.1}{6} = 0.01667 \]

### Step 5: Identify the limiting reagent
The limiting reagent is the one with the smallest ratio:
- TiO₂: 0.01792
- C: 0.12
- Cl₂: 0.01667

Thus, Cl₂ is the limiting reagent.

### Step 6: Calculate the amount of TiCl₄ produced
From the balanced equation, 6 moles of Cl₂ produce 3 moles of TiCl₄. Therefore:
- Moles of TiCl₄ produced from 0.1 moles of Cl₂:
\[ \text{Moles of TiCl}_4 = 0.1 \times \frac{3}{6} = 0.05 \text{ moles} \]

### Step 7: Calculate the mass of TiCl₄ produced
- Molar mass of TiCl₄:
\[ \text{Molar mass of TiCl}_4 = 48 + (4 \times 35.5) = 48 + 142 = 190 \text{ g/mol} \]

- Mass of TiCl₄ produced:
\[ \text{Mass of TiCl}_4 = \text{Moles} \times \text{Molar mass} = 0.05 \text{ moles} \times 190 \text{ g/mol} = 9.5 \text{ g} \]

### Final Answer
The mass of TiCl₄ that can be produced is **9.5 grams**.

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