The heat of combustion of sucrose `(C_(12)H_(22)O_(11))` is `1350 kcal//mol`. How much of heat will be liberated when `17.1 g` of sucrose is burnt ?

Options

A. `67.5 kcal`
B. `13.5 kcal`
C. `40.5 kcal`
D. `25.5 kcal`

Answer

To solve the problem of how much heat will be liberated when 17.1 g of sucrose is burnt, we can follow these steps:

### Step 1: Determine the molar mass of sucrose
The chemical formula for sucrose is \( C_{12}H_{22}O_{11} \). We need to calculate its molar mass.

- Carbon (C): \( 12 \times 12 \, \text{g/mol} = 144 \, \text{g/mol} \)
- Hydrogen (H): \( 22 \times 1 \, \text{g/mol} = 22 \, \text{g/mol} \)
- Oxygen (O): \( 11 \times 16 \, \text{g/mol} = 176 \, \text{g/mol} \)

Adding these together:
\[ \text{Molar mass of sucrose} = 144 + 22 + 176 = 342 \, \text{g/mol} \]

### Step 2: Use the heat of combustion
The heat of combustion of sucrose is given as \( 1350 \, \text{kcal/mol} \). This means that burning one mole (342 g) of sucrose releases 1350 kcal of heat.

### Step 3: Calculate the number of moles in 17.1 g of sucrose
To find out how many moles are in 17.1 g of sucrose, we can use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
Substituting the values:
\[ \text{Number of moles} = \frac{17.1 \, \text{g}}{342 \, \text{g/mol}} \approx 0.0500 \, \text{mol} \]

### Step 4: Calculate the heat released
Now, we can calculate the heat released when 17.1 g of sucrose is burnt:
\[ \text{Heat released} = \text{Number of moles} \times \text{Heat of combustion} \]
Substituting the values:
\[ \text{Heat released} = 0.0500 \, \text{mol} \times 1350 \, \text{kcal/mol} \approx 67.5 \, \text{kcal} \]

### Final Answer
The heat liberated when 17.1 g of sucrose is burnt is approximately **67.5 kcal**.

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