Combustion of glucose takes place according to the equation `C_(6)H_(12)O_(6)+6O_(2)` How much `rarr 6CO_(2)+6H_(2)O, DeltaH=-"72 kcal/mol"`.
energy will be released by the combustion of 1.6 g of glucose (Molecular mass of glucose = 180 g/mol)?

Options

A. 0.064 kcal
B. 0.64 kcal
C. 6.4 kcal
D. 64 kcal

Answer

To solve the problem of how much energy will be released by the combustion of 1.6 g of glucose, we can follow these steps:

### Step 1: Write down the combustion reaction
The combustion of glucose can be represented by the following balanced chemical equation:
\[ C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O \]
The enthalpy change (\( \Delta H \)) for this reaction is given as \( -72 \) kcal/mol.

### Step 2: Calculate the number of moles of glucose
To find out how many moles of glucose are present in 1.6 g, we can use the formula:
\[ n = \frac{W}{M} \]
Where:
- \( n \) = number of moles
- \( W \) = given mass of glucose (1.6 g)
- \( M \) = molar mass of glucose (180 g/mol)

Now, substituting the values:
\[ n = \frac{1.6 \, \text{g}}{180 \, \text{g/mol}} \]
\[ n = 0.008888 \, \text{mol} \]
Rounding this, we can say:
\[ n \approx 0.0089 \, \text{mol} \]

### Step 3: Calculate the energy released
Since the enthalpy change for the combustion of 1 mole of glucose is \( -72 \) kcal, we can find the energy released for \( 0.0089 \) moles:
\[ \text{Energy released} = n \times \Delta H \]
\[ \text{Energy released} = 0.0089 \, \text{mol} \times (-72 \, \text{kcal/mol}) \]
\[ \text{Energy released} = -0.6408 \, \text{kcal} \]
Rounding this, we can say:
\[ \text{Energy released} \approx -0.64 \, \text{kcal} \]

### Final Answer
The energy released by the combustion of 1.6 g of glucose is approximately \( 0.64 \) kcal.

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