Assume that a human body requires 2500 kcal of energy each day for metabolic activity and sucrose is the only source of energy, as per the equation
`C_(12)H_(22)O_(11)(S)+12 O_(2) (g) rarr 12 CO_(2)(g) + 11 H_(2) O (l) , Delta H = -5.6xx10^(6)J`.
(Fill up the blanks)
(a) The energy requirement of the human body per day is _______ kJ.
(b) The mass of sucrose required to provide this energy is ________ g and the volume of `CO_(2)` (at STP) produced is _______ litres .

Explanation

To solve the problem step by step, we will address each part of the question systematically.

### Given Information:
- Daily energy requirement of a human body: 2500 kcal
- Reaction:
\[ C_{12}H_{22}O_{11}(s) + 12 O_{2}(g) \rightarrow 12 CO_{2}(g) + 11 H_{2}O(l) \]
- Energy released per reaction: \(\Delta H = -5.6 \times 10^{6} \, \text{J}\)

### Part (a): Convert energy requirement from kcal to kJ
1. **Convert kcal to kJ**:
- We know that \(1 \, \text{kcal} = 4.184 \, \text{kJ}\).
- Therefore, to convert 2500 kcal to kJ:
\[ \text{Energy in kJ} = 2500 \, \text{kcal} \times 4.184 \, \text{kJ/kcal} \]
\[ = 10460 \, \text{kJ} \]

### Answer for Part (a):
The energy requirement of the human body per day is **10460 kJ**.

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### Part (b): Calculate the mass of sucrose required and the volume of CO₂ produced
1. **Energy released by sucrose**:
- The reaction releases \(5.6 \times 10^{6} \, \text{J}\) or \(5600 \, \text{kJ}\) for 342 g of sucrose.

2. **Calculate the mass of sucrose needed for 10460 kJ**:
- Set up a proportion based on the energy released:
\[ \frac{342 \, \text{g}}{5600 \, \text{kJ}} = \frac{x \, \text{g}}{10460 \, \text{kJ}} \]
- Cross-multiply to solve for \(x\):
\[ x = \frac{342 \, \text{g} \times 10460 \, \text{kJ}}{5600 \, \text{kJ}} \]
\[ = \frac{3574920}{5600} \approx 638 \, \text{g} \]

### Answer for the mass of sucrose:
The mass of sucrose required to provide this energy is **638 g**.

3. **Calculate the volume of CO₂ produced**:
- The reaction produces 12 moles of CO₂ for every 342 g of sucrose.
- At STP, 1 mole of gas occupies 22.4 L.
- Therefore, the volume of CO₂ produced from 342 g of sucrose is:
\[ \text{Volume of CO}_2 = 12 \, \text{moles} \times 22.4 \, \text{L/mole} = 268.8 \, \text{L} \]
- Now, we need to find the volume produced from 638 g of sucrose:
\[ \text{Volume of CO}_2 = \frac{268.8 \, \text{L}}{342 \, \text{g}} \times 638 \, \text{g} \]
\[ = \frac{268.8 \times 638}{342} \approx 501 \, \text{L} \]

### Answer for the volume of CO₂:
The volume of CO₂ produced is **501 L**.

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### Summary of Answers:
- (a) The energy requirement of the human body per day is **10460 kJ**.
- (b) The mass of sucrose required to provide this energy is **638 g** and the volume of CO₂ produced is **501 L**.

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