10 g carbon reacts with 100 g `Cl_(2)` to form `C Cl_(4)` The correct statement is

Options

A. Carbon is the limiting reagent
B. `Cl_(2)` is the limiting reagent
C. 108 . 41 g `C Cl_(4)` is formed
D. 0.833 moles of `C Cl_(4)` are formed

Answer

To solve the problem, we need to analyze the reaction between carbon and chlorine gas to form carbon tetrachloride (CCl₄). The balanced chemical equation for the reaction is:

\[ \text{C} + 2 \text{Cl}_2 \rightarrow \text{CCl}_4 \]

### Step 1: Calculate the moles of carbon (C)
Given mass of carbon = 10 g
Molar mass of carbon = 12 g/mol

Using the formula for moles:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]

Calculating moles of carbon:
\[ \text{Moles of C} = \frac{10 \, \text{g}}{12 \, \text{g/mol}} = 0.833 \, \text{moles} \]

### Step 2: Calculate the moles of chlorine (Cl₂)
Given mass of chlorine = 100 g
Molar mass of Cl₂ = 2 × 35.5 g/mol = 71 g/mol

Calculating moles of chlorine:
\[ \text{Moles of Cl}_2 = \frac{100 \, \text{g}}{71 \, \text{g/mol}} \approx 1.408 \, \text{moles} \]

### Step 3: Determine the limiting reagent
From the balanced equation, 1 mole of carbon reacts with 2 moles of chlorine. Therefore, for 0.833 moles of carbon, the required moles of chlorine would be:
\[ \text{Required moles of Cl}_2 = 2 \times 0.833 = 1.666 \, \text{moles} \]

We have only 1.408 moles of Cl₂ available, which is less than the required 1.666 moles. Thus, chlorine (Cl₂) is the limiting reagent.

### Step 4: Calculate the moles of CCl₄ produced
According to the reaction, 2 moles of Cl₂ produce 1 mole of CCl₄. Therefore, the moles of CCl₄ produced from the available Cl₂ can be calculated as:
\[ \text{Moles of CCl}_4 = \frac{1.408 \, \text{moles of Cl}_2}{2} = 0.704 \, \text{moles} \]

### Step 5: Calculate the mass of CCl₄ produced
Molar mass of CCl₄ = 12 g/mol (C) + 4 × 35.5 g/mol (Cl) = 12 + 142 = 154 g/mol

Now, calculating the mass of CCl₄ produced:
\[ \text{Mass of CCl}_4 = \text{Moles of CCl}_4 \times \text{Molar mass of CCl}_4 \]
\[ \text{Mass of CCl}_4 = 0.704 \, \text{moles} \times 154 \, \text{g/mol} = 108.416 \, \text{g} \approx 108.41 \, \text{g} \]

### Conclusion:
The correct statements from the options given are:
1. Chlorine (Cl₂) is the limiting reagent.
2. 108.41 g of CCl₄ is formed.