2.0 g of a sample contains mixture of `SiO_(2)` and `Fe_(2)O_(3)`. On very strong heating , it leaves a residue weighing `1.96 g`. The reaction responsible for loss of mass is given below .
`Fe_(2)O_(3)(s) to Fe_(3)O_(4)(s) + O_(2)(g)`, (unbalance equation) . What is the percentage by mass of `SiO_(2)` in original sample ?

Options

A. `10%`
B. `20%`
C. `40%`
D. `60%`

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the mass loss during heating.
The initial mass of the sample is given as 2.0 g, and the mass after heating is 1.96 g.

**Calculation:**
\[ \text{Mass loss} = \text{Initial mass} - \text{Final mass} = 2.0 \, \text{g} - 1.96 \, \text{g} = 0.04 \, \text{g} \]

### Step 2: Understand the reaction and balance it.
The reaction given is:
\[ \text{Fe}_2\text{O}_3(s) \rightarrow \text{Fe}_3\text{O}_4(s) + \text{O}_2(g) \]
To balance this reaction, we need to find the stoichiometric coefficients. The balanced equation is:
\[ 3 \text{Fe}_2\text{O}_3(s) \rightarrow 2 \text{Fe}_3\text{O}_4(s) + \frac{1}{2} \text{O}_2(g) \]

### Step 3: Relate mass loss to moles of oxygen evolved.
From the balanced equation, we see that 3 moles of \(\text{Fe}_2\text{O}_3\) produce \(\frac{1}{2}\) mole of \(\text{O}_2\). Therefore, for every mole of \(\text{O}_2\) produced, 6 moles of \(\text{Fe}_2\text{O}_3\) are consumed.

### Step 4: Calculate the number of moles of oxygen lost.
The molar mass of oxygen (\(\text{O}_2\)) is 32 g/mol.

**Calculation:**
\[ \text{Number of moles of } \text{O}_2 = \frac{\text{mass loss}}{\text{molar mass}} = \frac{0.04 \, \text{g}}{32 \, \text{g/mol}} = 0.00125 \, \text{mol} \]

### Step 5: Calculate the moles of \(\text{Fe}_2\text{O}_3\) that were present in the sample.
From the balanced equation, we know that:
\[ 1 \, \text{mol of } \text{O}_2 \text{ is produced from } 6 \, \text{mol of } \text{Fe}_2\text{O}_3 \]
Thus, for 0.00125 moles of \(\text{O}_2\):
\[ \text{Moles of } \text{Fe}_2\text{O}_3 = 6 \times 0.00125 = 0.0075 \, \text{mol} \]

### Step 6: Calculate the mass of \(\text{Fe}_2\text{O}_3\).
The molar mass of \(\text{Fe}_2\text{O}_3\) is 160 g/mol.

**Calculation:**
\[ \text{Mass of } \text{Fe}_2\text{O}_3 = \text{moles} \times \text{molar mass} = 0.0075 \, \text{mol} \times 160 \, \text{g/mol} = 1.2 \, \text{g} \]

### Step 7: Calculate the mass of \(\text{SiO}_2\) in the sample.
The total mass of the sample is 2.0 g, and the mass of \(\text{Fe}_2\text{O}_3\) is 1.2 g.

**Calculation:**
\[ \text{Mass of } \text{SiO}_2 = \text{Total mass} - \text{Mass of } \text{Fe}_2\text{O}_3 = 2.0 \, \text{g} - 1.2 \, \text{g} = 0.8 \, \text{g} \]

### Step 8: Calculate the percentage by mass of \(\text{SiO}_2\).
**Calculation:**
\[ \text{Percentage by mass of } \text{SiO}_2 = \left( \frac{\text{Mass of } \text{SiO}_2}{\text{Total mass}} \right) \times 100 = \left( \frac{0.8 \, \text{g}}{2.0 \, \text{g}} \right) \times 100 = 40\% \]

### Final Answer:
The percentage by mass of \(\text{SiO}_2\) in the original sample is **40%**.

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