The heat change during the reaction 24g C and 128g S following he change `C+S_(2)rarr CS_(2) , Delta H=22K` cal

Options

A. 22 K cal
B. 11 K cal
C. 44 K cal
D. 32 K cal

Explanation

To find the heat change during the reaction of 24 g of carbon with 128 g of sulfur, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction can be represented as:
\[ C + S_2 \rightarrow CS_2 \]

### Step 2: Determine the molar masses
- Molar mass of carbon (C) = 12 g/mol
- Molar mass of sulfur (S) = 32 g/mol (since sulfur exists as \( S_2 \), we need 2 moles for the reaction, so 2 x 32 g/mol = 64 g/mol for \( S_2 \))

### Step 3: Identify the stoichiometry of the reaction
From the balanced equation, 1 mole of carbon reacts with 1 mole of sulfur (which is 64 g of \( S_2 \)) to produce 1 mole of carbon disulfide (CS2).

### Step 4: Calculate the moles of reactants used
- Moles of carbon in 24 g:
\[ \text{Moles of } C = \frac{24 \text{ g}}{12 \text{ g/mol}} = 2 \text{ moles} \]
- Moles of sulfur in 128 g:
\[ \text{Moles of } S_2 = \frac{128 \text{ g}}{64 \text{ g/mol}} = 2 \text{ moles} \]

### Step 5: Relate the moles to the enthalpy change
The given enthalpy change (\( \Delta H \)) for the reaction is 22 kcal for the reaction of 1 mole of carbon with 1 mole of sulfur. Since we are using 2 moles of each reactant, the enthalpy change will also double.

### Step 6: Calculate the total heat change
\[ \Delta H = 2 \times 22 \text{ kcal} = 44 \text{ kcal} \]

### Final Answer
The heat change during the reaction of 24 g of carbon with 128 g of sulfur is **44 kcal**.

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