Aerobic respiration produces 264 gm of `CO_(2)` per 180 gm of glucose. The amount of `CO_(2)` produced for the same wieght of glucose in alcoholic fermentation shall be

Options

A. 132 gm
B. 528 gm
C. 88 gm
D. Zero

Answer

To solve the problem of how much CO₂ is produced during alcoholic fermentation for the same weight of glucose (180 gm), we can follow these steps:

### Step 1: Understand the process of aerobic respiration
In aerobic respiration, 180 gm of glucose produces 264 gm of CO₂. This is a key piece of information that we will use to compare with alcoholic fermentation.

### Step 2: Calculate the moles of CO₂ produced in aerobic respiration
The molecular weight of CO₂ is 44 gm/mol. To find the number of moles of CO₂ produced:
\[ \text{Moles of CO₂} = \frac{\text{grams of CO₂}}{\text{molecular weight of CO₂}} = \frac{264 \text{ gm}}{44 \text{ gm/mol}} = 6 \text{ moles} \]

### Step 3: Write the balanced equation for aerobic respiration
The balanced equation for aerobic respiration is:
\[ C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + \text{ATP} \]
From this equation, we see that 1 mole of glucose produces 6 moles of CO₂.

### Step 4: Understand the process of alcoholic fermentation
In alcoholic fermentation, glucose (C₆H₁₂O₆) is converted into ethanol (C₂H₅OH) and CO₂ in the absence of oxygen. The process can be summarized as:
\[ C_6H_{12}O_6 \rightarrow 2C_2H_5OH + 2CO_2 \]
This means that for every 1 mole of glucose, 2 moles of CO₂ are produced.

### Step 5: Calculate the grams of CO₂ produced in alcoholic fermentation
Since we have 180 gm of glucose, which is equivalent to 1 mole (molecular weight of glucose is 180 gm/mol), we can calculate the amount of CO₂ produced:
\[ \text{Moles of CO₂ from alcoholic fermentation} = 2 \text{ moles} \]
Now, converting moles of CO₂ to grams:
\[ \text{Grams of CO₂} = \text{Moles of CO₂} \times \text{molecular weight of CO₂} = 2 \text{ moles} \times 44 \text{ gm/mol} = 88 \text{ gm} \]

### Conclusion
Thus, the amount of CO₂ produced during alcoholic fermentation from 180 gm of glucose is **88 gm**.

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