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

Explanation

To solve the problem, we need to determine the weight of \( CO_2 \) produced when 170 Kcal of heat is evolved in the combustion of glucose. We start with the given reaction and the enthalpy change.

### Step-by-Step Solution:

1. **Identify the Reaction and Enthalpy Change**:
The combustion reaction of glucose is given as:
\[ C_6H_{12}O_6(s) + 9O_2(g) \rightarrow 6CO_2(g) + 6H_2O(g) \]
The enthalpy change (\( \Delta H \)) for this reaction is \(-680\) Kcal. This means that when 1 mole of glucose combusts, it releases 680 Kcal of heat and produces 6 moles of \( CO_2 \).

2. **Determine the Heat Released per Mole of \( CO_2 \)**:
From the reaction, we see that 6 moles of \( CO_2 \) are produced when 680 Kcal of heat is released. Therefore, the heat released per mole of \( CO_2 \) can be calculated as:
\[ \text{Heat per mole of } CO_2 = \frac{680 \text{ Kcal}}{6 \text{ moles}} = \frac{680}{6} \approx 113.33 \text{ Kcal/mole} \]

3. **Calculate Moles of \( CO_2 \) Produced from 170 Kcal**:
Now, we need to find out how many moles of \( CO_2 \) are produced when 170 Kcal of heat is released. We can set up a proportion:
\[ \text{Moles of } CO_2 = \frac{170 \text{ Kcal}}{113.33 \text{ Kcal/mole}} \approx 1.5 \text{ moles} \]

4. **Calculate the Mass of \( CO_2 \)**:
The molar mass of \( CO_2 \) is approximately 44 g/mole. Therefore, the mass of \( CO_2 \) produced can be calculated as:
\[ \text{Mass of } CO_2 = \text{Moles of } CO_2 \times \text{Molar mass of } CO_2 = 1.5 \text{ moles} \times 44 \text{ g/mole} = 66 \text{ g} \]

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