The heat of combustion of sucrose `(C_12H_22O_11)` is 1350 kcal/mol. How much of heat will be liberated when 161 g of sucrose is burnt ?

Options

A. 26.8 kcal
B. 57.63 kcal
C. 635.5 kcal
D. 52.87 kcal

Explanation

To solve the problem of how much heat will be liberated when 161 g of sucrose (C₁₂H₂₂O₁₁) is burnt, we can follow these steps:

### Step 1: Determine the molar mass of sucrose (C₁₂H₂₂O₁₁)
To find the molar mass, we sum the atomic masses of all the atoms in the sucrose molecule:
- Carbon (C): 12 atoms × 12 g/mol = 144 g/mol
- Hydrogen (H): 22 atoms × 1 g/mol = 22 g/mol
- Oxygen (O): 11 atoms × 16 g/mol = 176 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 kcal/mol. This means that when 1 mole (342 g) of sucrose is burnt, 1350 kcal of heat is released.

### Step 3: Calculate the amount of heat released for 161 g of sucrose
To find out how much heat is released when 161 g of sucrose is burnt, we first need to determine how many moles of sucrose are in 161 g.

Using the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
we can calculate:
\[ \text{Number of moles of sucrose} = \frac{161 \text{ g}}{342 \text{ g/mol}} \approx 0.470 \text{ moles} \]

### Step 4: Calculate the heat released
Now, we can find the heat released by multiplying the number of moles by the heat of combustion:
\[ \text{Heat released} = \text{Number of moles} \times \text{Heat of combustion} \]
\[ \text{Heat released} = 0.470 \text{ moles} \times 1350 \text{ kcal/mol} \approx 634.5 \text{ kcal} \]

### Conclusion
When 161 g of sucrose is burnt, approximately 634.5 kcal of heat will be liberated.