Which of the following `is//are` correct.
The following reaction occurs: ltrgt `CS_(2) + 3Cl_(2) overset(Delta)rarr C Cl_(4) + S_(2) Cl_(2)`
`1.0 g` of `CS_(2)` and `2.0 g` of `Cl_(2)` reacts.

Options

A. `0.714 g CS_(2)` is used in the reaction.
B. `0.286 g CS_(2)` is in formed.
C. `1.45 g of C Cl_(4)` is formed
D. `0.8 g Cl_(2)` is in excess

Explanation

To solve the problem step by step, we will first analyze the reaction and then calculate the moles of the reactants involved.

### Step 1: Write the balanced chemical equation
The given reaction is:
\[ \text{CS}_2 + 3\text{Cl}_2 \overset{\Delta}{\rightarrow} \text{CCl}_4 + \text{S}_2\text{Cl}_2 \]

### Step 2: Calculate the moles of CS₂
To find the moles of CS₂, we use the formula:
\[ \text{Moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]

The molar mass of CS₂ is calculated as follows:
- Carbon (C) = 12 g/mol
- Sulfur (S) = 32 g/mol

So, the molar mass of CS₂:
\[ 12 + 2 \times 32 = 12 + 64 = 76 \, \text{g/mol} \]

Now, substituting the values:
\[ \text{Moles of CS}_2 = \frac{1 \, \text{g}}{76 \, \text{g/mol}} = 0.013 \, \text{moles} \]

### Step 3: Calculate the moles of Cl₂
The molar mass of Cl₂ is:
- Chlorine (Cl) = 35.5 g/mol
- Therefore, the molar mass of Cl₂:
\[ 2 \times 35.5 = 71 \, \text{g/mol} \]

Now, substituting the values:
\[ \text{Moles of Cl}_2 = \frac{2 \, \text{g}}{71 \, \text{g/mol}} \approx 0.028 \, \text{moles} \]

### Step 4: Determine the limiting reagent
From the balanced equation, we see that 1 mole of CS₂ reacts with 3 moles of Cl₂. Therefore, for 0.013 moles of CS₂, the required moles of Cl₂ are:
\[ 0.013 \, \text{moles of CS}_2 \times 3 = 0.039 \, \text{moles of Cl}_2 \]

Since we only have 0.028 moles of Cl₂ available, Cl₂ is the limiting reagent.

### Step 5: Calculate the amount of CS₂ that reacts
Since Cl₂ is limiting, we will use all of it. The moles of CS₂ that will react with 0.028 moles of Cl₂ can be calculated as:
\[ \text{Moles of CS}_2 \text{ that reacts} = \frac{0.028}{3} \approx 0.00933 \, \text{moles} \]

### Step 6: Calculate the mass of CS₂ used
To find the mass of CS₂ that reacted:
\[ \text{Mass of CS}_2 \text{ used} = \text{Moles} \times \text{Molar mass} \]
\[ = 0.00933 \, \text{moles} \times 76 \, \text{g/mol} \approx 0.71 \, \text{g} \]

### Step 7: Calculate the remaining mass of CS₂
The initial mass of CS₂ was 1 g, so the remaining mass is:
\[ \text{Remaining mass of CS}_2 = 1 \, \text{g} - 0.71 \, \text{g} = 0.29 \, \text{g} \]

### Step 8: Calculate the amount of products formed
Since the reaction produces 1 mole of CCl₄ and 1 mole of S₂Cl₂ for every 3 moles of Cl₂, the moles of products formed will be equal to the moles of Cl₂ used:
\[ \text{Moles of CCl}_4 = \text{Moles of S}_2\text{Cl}_2 = 0.028 \, \text{moles} \]

### Step 9: Calculate the mass of CCl₄ formed
The molar mass of CCl₄ is:
\[ 12 + 4 \times 35.5 = 12 + 142 = 154 \, \text{g/mol} \]
\[ \text{Mass of CCl}_4 = 0.028 \, \text{moles} \times 154 \, \text{g/mol} \approx 4.32 \, \text{g} \]

### Conclusion
1. 0.71 g of CS₂ is used in the reaction.
2. 0.29 g of CS₂ is left.
3. 4.32 g of CCl₄ is formed.
4. Cl₂ is the limiting reagent and is completely used up.