1 gram of hydrated oxalic acid `[(COOH)_(2).2H_(2)O]` undergoes combustion it produces 2.2 kcal of heat. What will be its enthalpy of combustion `(DeltaH_(c))`?

Options

A. `-2.2 kcal`
B. `-126kcal`
C. `-277.3kcal`
D. `-423kcal`

Answer

To find the enthalpy of combustion (ΔHc) of hydrated oxalic acid \((COOH)_2 \cdot 2H_2O\), we can follow these steps:

### Step 1: Calculate the molar mass of hydrated oxalic acid \((COOH)_2 \cdot 2H_2O\)

1. **Carbon (C)**: There are 2 carbon atoms, each with an atomic mass of 12 g/mol.
\[ 2 \times 12 = 24 \text{ g/mol} \]

2. **Oxygen (O)**: There are 4 oxygen atoms from the oxalic acid and 2 from the water, totaling 6 oxygen atoms.
\[ 6 \times 16 = 96 \text{ g/mol} \]

3. **Hydrogen (H)**: There are 4 hydrogen atoms from the oxalic acid and 4 from the water, totaling 8 hydrogen atoms.
\[ 8 \times 1 = 8 \text{ g/mol} \]

4. **Total molar mass**:
\[ 24 + 96 + 8 = 128 \text{ g/mol} \]

### Step 2: Calculate the number of moles in 1 gram of hydrated oxalic acid

Using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
For 1 gram of hydrated oxalic acid:
\[ \text{Number of moles} = \frac{1 \text{ g}}{128 \text{ g/mol}} = 0.0078125 \text{ moles} \]

### Step 3: Calculate the enthalpy of combustion (ΔHc)

The enthalpy of combustion is given by the formula:
\[ \Delta H_c = -\frac{\text{heat liberated}}{\text{number of moles}} \]
Given that the heat liberated is 2.2 kcal:
\[ \Delta H_c = -\frac{2.2 \text{ kcal}}{0.0078125 \text{ moles}} = -281.6 \text{ kcal/mol} \]

### Step 4: Round the answer to the appropriate significant figures

The final answer can be rounded to:
\[ \Delta H_c \approx -281.6 \text{ kcal/mol} \]

### Summary of the Answer
The enthalpy of combustion of hydrated oxalic acid \((COOH)_2 \cdot 2H_2O\) is approximately \(-281.6 \text{ kcal/mol}\).

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