Carbon reacts with chlorine to form `"CC"l_(4).36`gm of carbon was mixed with 142 g of `Cl_(2)`. Calculate mass of `"CC"l_(4)` produced and the remaining mass of reactant .

Answer

To solve the problem of how much CCl₄ is produced and the remaining mass of reactants after the reaction between carbon and chlorine, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between carbon (C) and chlorine gas (Cl₂) to form carbon tetrachloride (CCl₄) is represented by the following balanced equation:

\[ C + 2Cl_2 \rightarrow CCl_4 \]

### Step 2: Calculate the molar masses
- Molar mass of Carbon (C) = 12 g/mol
- Molar mass of Chlorine (Cl₂) = 2 × 35.5 g/mol = 71 g/mol
- Molar mass of Carbon Tetrachloride (CCl₄) = 12 g/mol + 4 × 35.5 g/mol = 12 + 142 = 154 g/mol

### Step 3: Determine moles of reactants
Now, we calculate the number of moles of each reactant:
- Moles of Carbon (C):
\[ \text{Moles of C} = \frac{\text{Mass of C}}{\text{Molar mass of C}} = \frac{36 \text{ g}}{12 \text{ g/mol}} = 3 \text{ moles} \]

- Moles of Chlorine (Cl₂):
\[ \text{Moles of Cl₂} = \frac{\text{Mass of Cl₂}}{\text{Molar mass of Cl₂}} = \frac{142 \text{ g}}{71 \text{ g/mol}} = 2 \text{ moles} \]

### Step 4: Identify the limiting reagent
From the balanced equation, we see that 1 mole of C reacts with 2 moles of Cl₂. Therefore, to react with 2 moles of Cl₂, we need:
\[ \text{Moles of C required} = \frac{2 \text{ moles of Cl₂}}{2} = 1 \text{ mole of C} \]

We have 3 moles of C available, which means that Cl₂ is the limiting reagent because we only have enough Cl₂ to react with 1 mole of C.

### Step 5: Calculate the mass of CCl₄ produced
Since Cl₂ is the limiting reagent, we can calculate the mass of CCl₄ produced based on the moles of Cl₂:
- According to the balanced equation, 2 moles of Cl₂ produce 1 mole of CCl₄.
- Therefore, 2 moles of Cl₂ will produce 1 mole of CCl₄.

Since we have 2 moles of Cl₂, we will produce:
\[ \text{Moles of CCl₄ produced} = \frac{2 \text{ moles of Cl₂}}{2} = 1 \text{ mole of CCl₄} \]

Now, we can calculate the mass of CCl₄ produced:
\[ \text{Mass of CCl₄} = \text{Moles of CCl₄} \times \text{Molar mass of CCl₄} = 1 \text{ mole} \times 154 \text{ g/mol} = 154 \text{ g} \]

### Step 6: Calculate the remaining mass of reactants
- Since Cl₂ is the limiting reagent, it will be completely consumed:
\[ \text{Remaining mass of Cl₂} = 0 \text{ g} \]

- For Carbon, we initially had 3 moles (36 g), and we used 1 mole (12 g):
\[ \text{Remaining mass of C} = 36 \text{ g} - 12 \text{ g} = 24 \text{ g} \]

### Final Results
- Mass of CCl₄ produced = 154 g
- Remaining mass of Carbon = 24 g
- Remaining mass of Chlorine = 0 g