A sample of `NaClO_(3)` is converted by heat to `NaCl` with a loss of `0.16g`of oxygen. The residue is dissolved in water and precipitated as `AgCl`. The mass of `AgCl` (in g) obtained will be: (Given : Molar mass of `AgCl = 143.5 g mol^(-1)`)

Options

A. 0.54
B. 0.35
C. 0.48
D. 0.41

Explanation

To solve the problem step by step, we will follow the chemical reaction and stoichiometry principles.

### Step 1: Write the balanced chemical equation
The decomposition of sodium chlorate (\( \text{NaClO}_3 \)) can be represented as:
\[ 2 \text{NaClO}_3 \rightarrow 2 \text{NaCl} + 3 \text{O}_2 \]
This indicates that for every 2 moles of sodium chlorate, 2 moles of sodium chloride and 3 moles of oxygen are produced.

### Step 2: Calculate the moles of oxygen lost
Given that 0.16 g of oxygen is lost, we need to calculate the number of moles of oxygen (\( \text{O}_2 \)):
\[ \text{Moles of } O_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{0.16 \, \text{g}}{32 \, \text{g/mol}} = 0.005 \, \text{mol} \]

### Step 3: Determine the moles of sodium chloride produced
From the balanced equation, we know that 3 moles of oxygen correspond to 2 moles of sodium chloride. Thus, we can set up a ratio to find the moles of sodium chloride produced from the moles of oxygen:
\[ \text{Moles of } \text{NaCl} = \frac{2}{3} \times \text{Moles of } O_2 = \frac{2}{3} \times 0.005 \, \text{mol} = 0.00333 \, \text{mol} \]

### Step 4: Calculate the moles of silver chloride produced
The sodium chloride produced will react with silver ions to form silver chloride (\( \text{AgCl} \)). The reaction is:
\[ \text{NaCl} + \text{Ag}^+ \rightarrow \text{AgCl} + \text{Na}^+ \]
From the stoichiometry, 1 mole of sodium chloride produces 1 mole of silver chloride. Therefore, the moles of silver chloride produced will be equal to the moles of sodium chloride:
\[ \text{Moles of } \text{AgCl} = 0.00333 \, \text{mol} \]

### Step 5: Calculate the mass of silver chloride produced
To find the mass of silver chloride, we use the formula:
\[ \text{Mass} = \text{moles} \times \text{molar mass} \]
The molar mass of silver chloride (\( \text{AgCl} \)) is given as 143.5 g/mol. Thus:
\[ \text{Mass of } \text{AgCl} = 0.00333 \, \text{mol} \times 143.5 \, \text{g/mol} \approx 0.478 \, \text{g} \]

### Step 6: Round the result
Rounding this to two decimal places, we find that the mass of silver chloride produced is approximately:
\[ \text{Mass of } \text{AgCl} \approx 0.48 \, \text{g} \]

### Final Answer
The mass of \( \text{AgCl} \) obtained is **0.48 g**.

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