What is the equivalent weight of `C_(12)H_(22)O_(11)` in the following reaction?
`C_(12)H_(22)O_(11)+36HNO_(3) rarr 6H_(2)C_(2)O_(4)+36NO_(2)+23H_(2)O`

Options

A. `(342)/(36)`
B. `(342)/(12)`
C. `(342)/(22)`
D. `(342)/(3)`

Explanation

To find the equivalent weight of \( C_{12}H_{22}O_{11} \) in the given reaction:

\[ C_{12}H_{22}O_{11} + 36HNO_{3} \rightarrow 6H_{2}C_{2}O_{4} + 36NO_{2} + 23H_{2}O \]

we will follow these steps:

### Step 1: Determine the Molecular Weight of \( C_{12}H_{22}O_{11} \)

The molecular weight can be calculated using the atomic weights of carbon (C), hydrogen (H), and oxygen (O):

- Carbon (C): 12 g/mol
- Hydrogen (H): 1 g/mol
- Oxygen (O): 16 g/mol

Calculating the molecular weight:

\[ \text{Molecular Weight} = (12 \times 12) + (22 \times 1) + (11 \times 16) \]
\[ = 144 + 22 + 176 = 342 \text{ g/mol} \]

### Step 2: Determine the Change in Oxidation States

In this reaction, we need to identify how many electrons are transferred. We observe that the oxidation states of hydrogen and oxygen do not change, so we focus on carbon.

- In \( C_{12}H_{22}O_{11} \), the oxidation state of carbon is generally considered to be 0.
- In \( H_{2}C_{2}O_{4} \) (oxalic acid), the oxidation state of carbon is +3.

### Step 3: Calculate the Total Change in Electrons

Since \( C_{12}H_{22}O_{11} \) is converted to \( 6H_{2}C_{2}O_{4} \), we can calculate the total number of electrons transferred:

- Each carbon atom changes from 0 to +3, which means it loses 3 electrons.
- There are 12 carbon atoms in \( C_{12}H_{22}O_{11} \).

Thus, the total number of electrons lost is:

\[ \text{Total electrons} = 12 \text{ C atoms} \times 3 \text{ electrons/C atom} = 36 \text{ electrons} \]

### Step 4: Determine the Valence Factor

The valence factor (n) is the total number of electrons transferred per molecule of the substance. From our calculation, the valence factor for \( C_{12}H_{22}O_{11} \) is 36.

### Step 5: Calculate the Equivalent Weight

The equivalent weight is given by the formula:

\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{Valence Factor}} \]

Substituting the values we found:

\[ \text{Equivalent Weight} = \frac{342 \text{ g/mol}}{36} = 9.5 \text{ g/equiv} \]

### Final Answer

The equivalent weight of \( C_{12}H_{22}O_{11} \) in the reaction is \( 9.5 \text{ g/equiv} \).

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