Question

Fortification of food with iron is done using $\mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O}$. The mass in grams of the $\mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O}$ required to achieve 12 ppm of iron in 150 kg of wheat is _______ (Nearest integer).} (Given : Molar mass of $\mathrm{Fe}, \mathrm{S}$ and O respectively are 56,32 and $16 \mathrm{~g} \mathrm{~mol}^{-1}$ )

Hint:

Use the molar mass and stoichiometry to calculate the mass of the compound required.

Answer 1

Correct Answer: 9

To calculate the required mass of $\mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}$ for 12 ppm iron in 150 kg of wheat, proceed as follows:

Define ppm: 1 ppm equates to 1 mg of solute per 1 kg of solution. Thus, the necessary mass of iron for 12 ppm in 150 kg of wheat is:

\( \text{Mass of Fe} = 12 \,\text{mg/kg} \times 150 \,\text{kg} = 1800 \,\text{mg} = 1.8 \,\text{g} \)

Determine the molar mass of $\mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}$:

Element	Atomic Mass (g/mol)	Atoms	Total (g/mol)
Fe	56	1	56
S	32	1	32
O	16	11 (4+7)	176
H	1	14 (2x7)	14
Total	278 g/mol

Calculate the moles of $\mathrm{Fe}$ corresponding to 1.8 g:

\( \text{Moles of Fe} = \frac{1.8 \,\text{g}}{56 \,\text{g/mol}} \approx 0.03214 \,\text{mol} \)

Since each $\mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}$ molecule contains one iron atom, the moles of $\mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}$ are:

\( \text{Moles of } \mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O} = 0.03214 \,\text{mol} \)

Calculate the mass of $\mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}$:

\( \text{Mass} = 0.03214 \,\text{mol} \times 278 \,\text{g/mol} \approx 8.936 \,\text{g} \)

Rounded to the nearest integer: 9 g

Verification: The calculated value is 9 g, which falls within the range 9–9. The result is confirmed.

Therefore, 9 g of $\mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}$ is required.