What is the percentage loss in water when `BaCI_2. 2H_2 O` becomes completely anhydrous?

Options

A. 0.1475
B. 0.2822
C. 0.8525
D. 0.4327

Answer

To find the percentage loss in water when BaCl₂·2H₂O becomes completely anhydrous, we can follow these steps:

### Step 1: Write the decomposition reaction
When barium chloride dihydrate (BaCl₂·2H₂O) is heated, it loses water and becomes anhydrous barium chloride (BaCl₂). The reaction can be written as:
\[ \text{BaCl}_2 \cdot 2\text{H}_2\text{O} \rightarrow \text{BaCl}_2 + 2\text{H}_2\text{O} \]

### Step 2: Determine the moles of water produced
From the reaction, we see that 1 mole of BaCl₂·2H₂O produces 2 moles of water (H₂O).

### Step 3: Calculate the mass of water lost
To find the mass of the water lost, we need to calculate the molecular weight of water (H₂O):
- The atomic weight of Hydrogen (H) = 1.01 g/mol
- The atomic weight of Oxygen (O) = 16 g/mol

The molecular weight of H₂O is:
\[ \text{Molecular weight of H}_2\text{O} = 2 \times 1.01 + 16 = 18.02 \text{ g/mol} \]

Since we have 2 moles of water:
\[ \text{Mass of water lost} = 2 \times 18.02 = 36.04 \text{ g} \]

### Step 4: Calculate the molecular weight of BaCl₂·2H₂O
Next, we calculate the molecular weight of BaCl₂·2H₂O:
- Atomic weight of Barium (Ba) = 137.33 g/mol
- Atomic weight of Chlorine (Cl) = 35.45 g/mol

The molecular weight of BaCl₂·2H₂O is:
\[ \text{Molecular weight of BaCl}_2 \cdot 2\text{H}_2\text{O} = 137.33 + 2 \times 35.45 + 2 \times 18.02 \]
\[ = 137.33 + 70.90 + 36.04 = 244.27 \text{ g/mol} \]

### Step 5: Calculate the percentage loss of water
Now we can calculate the percentage loss of water:
\[ \text{Percentage loss of water} = \left( \frac{\text{Mass of water lost}}{\text{Molecular weight of BaCl}_2 \cdot 2\text{H}_2\text{O}} \right) \times 100 \]
\[ = \left( \frac{36.04}{244.27} \right) \times 100 \approx 14.75\% \]

Thus, the percentage loss in water when BaCl₂·2H₂O becomes completely anhydrous is approximately **14.75%**.