How many milli gram of iron `(Fe^(2+))` are equal to 1 mL of `0.1055NK_2Cr_2O_7` equivalent?

Options

A. 5.9 mg
B. 0.59 mg
C. 59 mg
D. `5.9 xx 10^(-3) mg`

Answer

To find out how many milligrams of iron `(Fe^(2+))` are equal to 1 mL of `0.1055N K2Cr2O7` equivalent, we can follow these steps:

### Step 1: Understand the Concept of Equivalents
Equivalents are used to express the amount of a substance that reacts with or supplies one mole of hydrogen ions (H⁺) in a reaction. For this problem, we need to equate the equivalents of `Fe^(2+)` and `K2Cr2O7`.

### Step 2: Write the Equation for Equivalents
The equivalents of `Fe^(2+)` will be equal to the equivalents of `K2Cr2O7`:
\[ \text{Equivalent of } Fe^{2+} = \text{Equivalent of } K2Cr2O7 \]

### Step 3: Calculate the Equivalent of K2Cr2O7
The equivalent of a solution can be calculated using the formula:
\[ \text{Equivalent} = \text{Normality} \times \text{Volume (in L)} \]
Since we have 1 mL of `0.1055N K2Cr2O7`, we convert the volume to liters:
\[ 1 \text{ mL} = 0.001 \text{ L} \]
Thus,
\[ \text{Equivalent of } K2Cr2O7 = 0.1055 \, N \times 0.001 \, L = 0.0001055 \, \text{equivalents} \]

### Step 4: Relate the Charge of Iron
For `Fe^(2+)`, the charge is 2 (since it can lose 2 electrons). Therefore, the number of equivalents for `Fe^(2+)` is given by:
\[ \text{Equivalent of } Fe^{2+} = \frac{\text{mass of Fe}}{\text{molar mass of Fe}} \]
Where the molar mass of iron (Fe) is 56 g/mol.

### Step 5: Set Up the Equation
Since the equivalents of `Fe^(2+)` equal the equivalents of `K2Cr2O7`, we can set up the equation:
\[ \frac{m}{56} = 0.0001055 \]
Where \( m \) is the mass of iron in grams.

### Step 6: Solve for Mass of Iron
Rearranging the equation gives:
\[ m = 0.0001055 \times 56 \]
Calculating this gives:
\[ m = 0.005918 \, \text{grams} \]

### Step 7: Convert Grams to Milligrams
To convert grams to milligrams, we multiply by 1000:
\[ m = 0.005918 \, \text{grams} \times 1000 = 5.918 \, \text{milligrams} \]

### Final Answer
Thus, the mass of iron `(Fe^(2+))` that is equal to 1 mL of `0.1055N K2Cr2O7` is approximately **5.918 mg**.