Ferric sulphate on heating gives sulphur trioxide. The ratio between the weights of oxygen and sulphur present in `SO_(3)` obtained by heating 1 kg of ferric sulphate is

Options

A. `2:3`
B. `1:3`
C. `3:1`
D. `3:2`

Explanation

To find the ratio between the weights of oxygen and sulfur present in \( SO_3 \) obtained by heating 1 kg of ferric sulfate, we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced equation for the thermal decomposition of ferric sulfate (\( Fe_2(SO_4)_3 \)) is:
\[ Fe_2(SO_4)_3 \rightarrow Fe_2O_3 + 3 SO_3 \]
This indicates that 1 mole of ferric sulfate produces 3 moles of sulfur trioxide (\( SO_3 \)).

### Step 2: Calculate the molar mass of ferric sulfate
The molar mass of \( Fe_2(SO_4)_3 \) can be calculated as follows:
- Atomic mass of Fe = 56 g/mol (2 Fe = \( 2 \times 56 = 112 \) g)
- Atomic mass of S = 32 g/mol (3 S = \( 3 \times 32 = 96 \) g)
- Atomic mass of O = 16 g/mol (12 O = \( 12 \times 16 = 192 \) g)

Adding these together:
\[ Molar \ mass \ of \ Fe_2(SO_4)_3 = 112 + 96 + 192 = 400 \ g/mol \]

### Step 3: Determine the amount of \( SO_3 \) produced from 1 kg of \( Fe_2(SO_4)_3 \)
Since 1 mole of \( Fe_2(SO_4)_3 \) produces 3 moles of \( SO_3 \), we can find out how much \( SO_3 \) is produced from 1 kg (1000 g) of \( Fe_2(SO_4)_3 \):
\[ \text{Moles of } Fe_2(SO_4)_3 = \frac{1000 \ g}{400 \ g/mol} = 2.5 \ moles \]
Thus, the moles of \( SO_3 \) produced:
\[ \text{Moles of } SO_3 = 2.5 \ moles \times 3 = 7.5 \ moles \]

### Step 4: Calculate the mass of \( SO_3 \)
The molar mass of \( SO_3 \) is:
\[ Molar \ mass \ of \ SO_3 = 32 \ g/mol \ (S) + 3 \times 16 \ g/mol \ (O) = 32 + 48 = 80 \ g/mol \]
Now, calculate the mass of \( SO_3 \):
\[ \text{Mass of } SO_3 = 7.5 \ moles \times 80 \ g/mol = 600 \ g \]

### Step 5: Find the mass of sulfur and oxygen in \( SO_3 \)
In 80 g of \( SO_3 \):
- Mass of sulfur = 32 g
- Mass of oxygen = 48 g

Now, we can find the mass of sulfur and oxygen in 600 g of \( SO_3 \):
- Mass of sulfur in 600 g of \( SO_3 \):
\[ \text{Mass of S} = \frac{32 \ g}{80 \ g} \times 600 \ g = 240 \ g \]
- Mass of oxygen in 600 g of \( SO_3 \):
\[ \text{Mass of O} = \frac{48 \ g}{80 \ g} \times 600 \ g = 360 \ g \]

### Step 6: Calculate the ratio of the weights of oxygen to sulfur
Now, we can find the ratio of the weights of oxygen to sulfur:
\[ \text{Ratio} = \frac{\text{Mass of O}}{\text{Mass of S}} = \frac{360 \ g}{240 \ g} = \frac{3}{2} \]

### Final Answer:
The ratio between the weights of oxygen and sulfur present in \( SO_3 \) obtained by heating 1 kg of ferric sulfate is \( 3:2 \).

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