`K_(2)Cr_(2)O_(7)` is obtained in the following steps:
`2FeCrO_(4)+2Na_(2)CO_(3)+OtoFe_(2)O_(3)+2Na_(2)CrO_(4)+2CO_(2)`
`2Na_(2)CrO_(4)+H_(2)SO_(4)toNa_(2)Cr_(2)O_(7)+H_(2)O+Na_(2)SO_(4)`
`Na_(2)Cr_(2)O_(7)+2KCltoK_(2)Cr_(2)+2NaCl`
To get 0.25 " mol of "`K_(2)Cr_(2)O_(7)`, " mol of "`50%` pure `FeCrO_(4)` required.

Options

A. 1 mol
B. 0.50 mol
C. 0.25 mol
D. 0.125 mol

Explanation

To determine how much 50% pure `FeCrO_(4)` is required to produce `0.25` moles of `K_(2)Cr_(2)O_(7)`, we can follow these steps:

### Step 1: Understand the stoichiometry of the reactions
From the provided reactions, we can see that:
- From the first reaction, `2FeCrO_(4)` produces `1K_(2)Cr_(2)O_(7)`.
This means that `2 moles of FeCrO_(4)` are required to produce `1 mole of K_(2)Cr_(2)O_(7)`.

### Step 2: Calculate the moles of `FeCrO_(4)` needed for `0.25` moles of `K_(2)Cr_(2)O_(7)`
Since `2 moles of FeCrO_(4)` yield `1 mole of K_(2)Cr_(2)O_(7)`, we can set up a proportion:
- If `1 mole of K_(2)Cr_(2)O_(7)` requires `2 moles of FeCrO_(4)`, then `0.25 moles of K_(2)Cr_(2)O_(7)` will require:

\[ \text{Moles of } FeCrO_(4) = 0.25 \text{ moles } K_(2)Cr_(2)O_(7) \times \frac{2 \text{ moles } FeCrO_(4)}{1 \text{ mole } K_(2)Cr_(2)O_(7)} = 0.50 \text{ moles of } FeCrO_(4) \]

### Step 3: Adjust for the purity of `FeCrO_(4)`
Since the `FeCrO_(4)` is only `50%` pure, we need to calculate the amount of the impure `FeCrO_(4)` required to obtain `0.50` moles of pure `FeCrO_(4)`.

Let \( x \) be the amount of `50%` pure `FeCrO_(4)` required. The amount of pure `FeCrO_(4)` in \( x \) moles is given by:

\[ \text{Pure } FeCrO_(4) = 0.50 \times x \]

We need this to equal `0.50` moles of pure `FeCrO_(4)`:

\[ 0.50 \times x = 0.50 \]

### Step 4: Solve for \( x \)
To find \( x \):

\[ x = \frac{0.50}{0.50} = 1 \text{ mole of } 50\% \text{ pure } FeCrO_(4) \]

### Conclusion
To produce `0.25` moles of `K_(2)Cr_(2)O_(7)`, you will need `1 mole` of `50%` pure `FeCrO_(4)`.

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