A person inhales `640g ` of `O_(20` per day. If all the `O_(2)` is used for converting sugar into `CO_(2)` and `H_(2)O`, how much sucrose `(C_(12)H_(22)O_(11))` is consumed in the body in one day and what is the heat evolved ?
`(DeltaH_("combination of sucrose ")=-5645kJ mol^(-1))`

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the number of moles of oxygen inhaled per day.
Given:
- Mass of \( O_2 \) inhaled = 640 g
- Molar mass of \( O_2 \) = 32 g/mol (since \( O \) has a molar mass of 16 g/mol, and \( O_2 \) has two oxygen atoms)

\[ \text{Number of moles of } O_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{640 \, \text{g}}{32 \, \text{g/mol}} = 20 \, \text{moles} \]

### Step 2: Write the balanced chemical equation for the combustion of sucrose.
The balanced equation for the combustion of sucrose \( C_{12}H_{22}O_{11} \) is:

\[ C_{12}H_{22}O_{11} + 12 O_2 \rightarrow 12 CO_2 + 11 H_2O \]

From the balanced equation, we see that 1 mole of sucrose reacts with 12 moles of \( O_2 \).

### Step 3: Calculate the number of moles of sucrose consumed.
Using the stoichiometry from the balanced equation:

\[ \text{If } 12 \text{ moles of } O_2 \text{ are required for } 1 \text{ mole of sucrose, then } 20 \text{ moles of } O_2 \text{ will require:} \]

\[ \text{Number of moles of sucrose} = \frac{20 \, \text{moles of } O_2}{12} = \frac{20}{12} \approx 1.67 \, \text{moles of sucrose} \]

### Step 4: Calculate the mass of sucrose consumed.
The molar mass of sucrose \( C_{12}H_{22}O_{11} \) is calculated as follows:

\[ \text{Molar mass of sucrose} = (12 \times 12) + (22 \times 1) + (11 \times 16) = 144 + 22 + 176 = 342 \, \text{g/mol} \]

Now, calculate the mass of sucrose consumed:

\[ \text{Mass of sucrose} = \text{Number of moles} \times \text{Molar mass} = 1.67 \, \text{moles} \times 342 \, \text{g/mol} \approx 570 \, \text{g} \]

### Step 5: Calculate the heat evolved.
Given that the enthalpy of combustion of sucrose is \( \Delta H = -5645 \, \text{kJ/mol} \), we can calculate the heat evolved for the amount of sucrose consumed:

\[ \text{Heat evolved} = \text{Number of moles of sucrose} \times \Delta H \]

\[ \text{Heat evolved} = 1.67 \, \text{moles} \times (-5645 \, \text{kJ/mol}) \approx -9408.33 \, \text{kJ} \]

### Final Answers
- Mass of sucrose consumed per day: **570 g**
- Heat evolved: **-9408.33 kJ**

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