If `C_(6)H_(12)O_(6)(s)+9O_(2)(g)rarr6CO_(2)(g)+6H_(2)O(g) , Delta H=-680` Kcal The weight of `CO_(2)(g)` produced when 170 Kcal of heat is evolved in the combustion of glucose is :-

Options

A. 265 gm
B. 66 gm
C. 11 gm
D. 64 gm

Answer

To solve the problem, we need to determine the weight of \( CO_2 \) produced when 170 Kcal of heat is evolved during the combustion of glucose. We know from the reaction that the enthalpy change (\( \Delta H \)) for the combustion of glucose is -680 Kcal, which corresponds to the production of 6 moles of \( CO_2 \).

### Step-by-Step Solution:

1. **Identify the relationship between heat evolved and moles of \( CO_2 \)**:
The reaction states that for the combustion of glucose, -680 Kcal of heat is released when 6 moles of \( CO_2 \) are produced. This means that:
\[ 680 \text{ Kcal} \rightarrow 6 \text{ moles of } CO_2 \]

2. **Calculate the amount of \( CO_2 \) produced for 170 Kcal**:
We need to find out how many moles of \( CO_2 \) are produced when 170 Kcal of heat is evolved. We can set up a proportion:
\[ \frac{6 \text{ moles of } CO_2}{680 \text{ Kcal}} = \frac{x \text{ moles of } CO_2}{170 \text{ Kcal}} \]
Cross-multiplying gives:
\[ x = \frac{6 \times 170}{680} \]
Simplifying this:
\[ x = \frac{1020}{680} = 1.5 \text{ moles of } CO_2 \]

3. **Calculate the mass of \( CO_2 \)**:
Now, we need to convert moles of \( CO_2 \) to grams. The molar mass of \( CO_2 \) is calculated as follows:
\[ \text{Molar mass of } CO_2 = 12 \text{ (C)} + 16 \times 2 \text{ (O)} = 44 \text{ g/mol} \]
Therefore, the mass of \( CO_2 \) produced is:
\[ \text{Mass of } CO_2 = \text{moles} \times \text{molar mass} = 1.5 \text{ moles} \times 44 \text{ g/mol} = 66 \text{ g} \]

### Final Answer:
The weight of \( CO_2 \) produced when 170 Kcal of heat is evolved is **66 grams**.