In cellular respiration, from 180 g of glucose which of the following is formed ?

Options

A. `264g CO_2 + 190 g H_2O + 686 Kcal`
B. `264g CO_2 + 108 g H_2O + 686 Kcal`
C. `390g CO_2 + 108 g H_2O + 391 Kcal`
D. `390g CO_2 + 264g H_2O + 391 Kcal`

Answer

To solve the question regarding the products formed from 180 g of glucose during cellular respiration, we can follow these steps:

### Step 1: Write the Cellular Respiration Equation
The balanced equation for cellular respiration is:
\[ \text{C}_6\text{H}_{12}\text{O}_6 + 6 \text{O}_2 \rightarrow 6 \text{CO}_2 + 6 \text{H}_2\text{O} + 686 \text{ kcal} \]

### Step 2: Determine the Molar Mass of Glucose
The molar mass of glucose (C₆H₁₂O₆) can be calculated as follows:
- Carbon (C): 12 g/mol × 6 = 72 g/mol
- Hydrogen (H): 1 g/mol × 12 = 12 g/mol
- Oxygen (O): 16 g/mol × 6 = 96 g/mol
- Total molar mass of glucose = 72 + 12 + 96 = 180 g/mol

### Step 3: Calculate the Products from 180 g of Glucose
From the balanced equation, we see that 1 mole of glucose (180 g) produces:
- 6 moles of carbon dioxide (CO₂)
- 6 moles of water (H₂O)
- 686 kilocalories of energy

### Step 4: Calculate the Mass of Carbon Dioxide and Water Produced
1. **Carbon Dioxide (CO₂)**:
- Molar mass of CO₂ = 12 g/mol (C) + 16 g/mol × 2 (O) = 44 g/mol
- Mass of 6 moles of CO₂ = 6 × 44 g = 264 g

2. **Water (H₂O)**:
- Molar mass of H₂O = 1 g/mol (H) × 2 + 16 g/mol (O) = 18 g/mol
- Mass of 6 moles of H₂O = 6 × 18 g = 108 g

### Step 5: Summarize the Products
From 180 g of glucose, the products formed are:
- 264 g of carbon dioxide (CO₂)
- 108 g of water (H₂O)
- 686 kilocalories of energy

### Conclusion
The correct answer is:
- 264 g of carbon dioxide, 108 g of water, and 686 kilocalories of energy.

### Final Answer
The answer is option B: 264 g of carbon dioxide + 108 g of water + 686 kilocalories.

---