Calculate the heat required to make 6.4kg of `CaC_2` from `CaO(s) and C(s)` from the reaction that `DeltaH_f^@(CaO) = -151.6 kcal`
`DeltaH_f^@(CaC_2) = -14.2 kcal`
`DeltaH_f^@(CO) = -26.4 kcal`.

Options

A. `5624 kcal`
B. `86.24 xx 10^3 kcal`
C. `1.11 xx 10^4 kcal`
D. `1100 kcal`

Answer

To calculate the heat required to make 6.4 kg of calcium carbide (CaC₂) from calcium oxide (CaO) and carbon (C), we can follow these steps:

### Step 1: Write the Balanced Chemical Equation
The reaction can be represented as:
\[ \text{CaO(s)} + 3\text{C(s)} \rightarrow \text{CaC}_2\text{(s)} + \text{CO(g)} \]

### Step 2: Determine the Standard Heat of Formation Values
We are given the following standard heats of formation:
- \( \Delta H_f^{\circ}(\text{CaO}) = -151.6 \, \text{kcal} \)
- \( \Delta H_f^{\circ}(\text{CaC}_2) = -14.2 \, \text{kcal} \)
- \( \Delta H_f^{\circ}(\text{CO}) = -26.4 \, \text{kcal} \)

### Step 3: Calculate the Heat of Reaction
Using the formula for the heat of reaction:
\[ \Delta H = \sum \Delta H_f^{\circ}(\text{products}) - \sum \Delta H_f^{\circ}(\text{reactants}) \]

For the reaction:
- Products: \( \Delta H_f^{\circ}(\text{CaC}_2) + \Delta H_f^{\circ}(\text{CO}) \)
- Reactants: \( \Delta H_f^{\circ}(\text{CaO}) + 3 \Delta H_f^{\circ}(\text{C}) \)

Since the heat of formation of carbon (C) in its standard state is zero, we have:
\[ \Delta H = \left( -14.2 + (-26.4) \right) - \left( -151.6 \right) \]

Calculating this gives:
\[ \Delta H = (-14.2 - 26.4 + 151.6) \]
\[ \Delta H = 111 \, \text{kcal} \]

### Step 4: Determine the Molar Mass of Calcium Carbide (CaC₂)
The molar mass of CaC₂ is calculated as follows:
- Calcium (Ca): 40 g/mol
- Carbon (C): 12 g/mol
- Therefore, for CaC₂:
\[ 40 + (2 \times 12) = 40 + 24 = 64 \, \text{g/mol} \]

### Step 5: Calculate the Heat Required for 6.4 kg of CaC₂
We need to find out how much heat is required to produce 6.4 kg (6400 g) of CaC₂.

Using the proportion:
- Heat required for 64 g of CaC₂ = 111 kcal
- Heat required for 6400 g of CaC₂ = \( X \)

Setting up the equation:
\[ X = \frac{111 \, \text{kcal} \times 6400 \, \text{g}}{64 \, \text{g}} \]
\[ X = 111 \times 100 = 11100 \, \text{kcal} \]
\[ X = 1.11 \times 10^4 \, \text{kcal} \]

### Final Answer
The heat required to make 6.4 kg of calcium carbide is:
\[ \boxed{1.11 \times 10^4 \, \text{kcal}} \]

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