For the reaction, `2Fe(NO_3)+3Na_2COrarrFe_2(CO_3)_3+6NaNO_3` initially `2.5` mole of `Fe(NO_3)_3 and 3.6` mole of `Na_2 CO_3` are taken. If `6.3` mole of `NaNO_3` is obtained then % yield of given reaction is :

Options

A. 50
B. 84
C. 87.5
D. 100

Answer

To calculate the percent yield of the reaction given the initial moles of reactants and the amount of product formed, we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ 2 \text{Fe(NO}_3\text{)}_3 + 3 \text{Na}_2\text{CO}_3 \rightarrow \text{Fe}_2(\text{CO}_3)_3 + 6 \text{NaNO}_3 \]

### Step 2: Identify the initial amounts of reactants
We have:
- Initial moles of \(\text{Fe(NO}_3\text{)}_3 = 2.5\) moles
- Initial moles of \(\text{Na}_2\text{CO}_3 = 3.6\) moles

### Step 3: Determine the limiting reagent
From the balanced equation, we see that:
- 2 moles of \(\text{Fe(NO}_3\text{)}_3\) react with 3 moles of \(\text{Na}_2\text{CO}_3\).

To find out how much \(\text{Na}_2\text{CO}_3\) is needed for 2.5 moles of \(\text{Fe(NO}_3\text{)}_3\):
\[ \text{Moles of } \text{Na}_2\text{CO}_3 \text{ required} = \frac{3}{2} \times 2.5 = 3.75 \text{ moles} \]

Since we only have 3.6 moles of \(\text{Na}_2\text{CO}_3\), which is less than 3.75 moles, \(\text{Na}_2\text{CO}_3\) is the limiting reagent.

### Step 4: Calculate the theoretical yield of \(\text{NaNO}_3\)
From the balanced equation:
- 3 moles of \(\text{Na}_2\text{CO}_3\) produce 6 moles of \(\text{NaNO}_3\).

Using the amount of the limiting reagent:
\[ \text{Moles of } \text{NaNO}_3 \text{ produced} = \frac{6}{3} \times 3.6 = 7.2 \text{ moles} \]

### Step 5: Compare the theoretical yield with the actual yield
The actual yield of \(\text{NaNO}_3\) obtained is given as 6.3 moles.

### Step 6: Calculate the percent yield
Percent yield is calculated using the formula:
\[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \]
Substituting the values:
\[ \text{Percent Yield} = \left( \frac{6.3}{7.2} \right) \times 100 \approx 87.5\% \]

### Final Answer
The percent yield of the reaction is approximately **87.5%**.

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