Calculated the mass of `FeSO_(4).7H_(2)O`, which must be added in 100 kg of wheat to get 10 PPM of Fe.

Answer

To find the mass of `FeSO4.7H2O` that must be added to 100 kg of wheat to achieve a concentration of 10 PPM of iron (Fe), we can follow these steps:

### Step 1: Understand PPM
PPM (parts per million) is a way to express very dilute concentrations of substances. It is defined as:
\[ \text{PPM} = \frac{\text{mass of solute}}{\text{total mass of solution}} \times 10^6 \]
In this case, the solute is iron (Fe), and the total mass of the solution is the mass of wheat.

### Step 2: Set Up the Equation
Given that we want 10 PPM of Fe in 100 kg of wheat, we can set up the equation:
\[ 10 = \frac{\text{mass of Fe}}{100 \, \text{kg}} \times 10^6 \]

### Step 3: Convert Mass of Wheat to Grams
Since 1 kg = 1000 grams, we convert 100 kg of wheat to grams:
\[ 100 \, \text{kg} = 100 \times 1000 \, \text{g} = 100000 \, \text{g} \]

### Step 4: Solve for Mass of Fe
Now, substitute the mass of wheat into the PPM equation:
\[ 10 = \frac{\text{mass of Fe}}{100000 \, \text{g}} \times 10^6 \]
Rearranging gives:
\[ \text{mass of Fe} = \frac{10 \times 100000 \, \text{g}}{10^6} = 1 \, \text{g} \]

### Step 5: Calculate the Mass of `FeSO4.7H2O`
Next, we need to find out how much `FeSO4.7H2O` is required to provide 1 g of Fe.

1. **Molar Mass of `FeSO4.7H2O`:**
- Molar mass of Fe = 56 g/mol
- Molar mass of S = 32 g/mol
- Molar mass of O = 16 g/mol (4 O in SO4)
- Molar mass of H2O = 18 g/mol (7 H2O)

Therefore, the molar mass of `FeSO4.7H2O` is:
\[ 56 + 32 + (4 \times 16) + (7 \times 18) = 56 + 32 + 64 + 126 = 278 \, \text{g/mol} \]

2. **Calculate the Mass of `FeSO4.7H2O` Needed for 1 g of Fe:**
Using the ratio of the masses:
\[ \text{mass of } FeSO4.7H2O = \frac{278 \, \text{g}}{56 \, \text{g}} \times 1 \, \text{g} \approx 4.96 \, \text{g} \]

### Step 6: Round the Answer
Rounding 4.96 g gives approximately:
\[ \text{mass of } FeSO4.7H2O \approx 5 \, \text{g} \]

### Final Answer
The mass of `FeSO4.7H2O` that must be added to 100 kg of wheat to achieve 10 PPM of Fe is approximately **5 grams**.

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