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 determine the mass of $\mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}$ needed to achieve 12 ppm of iron in 150 kg of wheat, follow these steps:

Define ppm: 1 ppm is 1 mg of solute per 1 kg of solution. Therefore, for 12 ppm of iron in 150 kg of wheat, the mass of iron needed is:

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

To find the mass of $\mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}$, start by calculating its molar mass:

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 $\mathrm{Fe}$ in 1.8 g:

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

Convert moles of $\mathrm{Fe}$ to moles of $\mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}$ since each unit contains one Fe atom:

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

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

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

Round to the nearest integer: 9 g

Verify the solution is within the range 9–9. It is correct.

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

Answer 2

1. Calculate the mass of iron required: \[ \text{Mass of iron} = \frac{12 \mathrm{~ppm} \times 150 \mathrm{~kg}}{10^6} = 1.8 \mathrm{~g} \]
2. Calculate the moles of iron: \[ \text{Moles of iron} = \frac{1.8 \mathrm{~g}}{56 \mathrm{~g/mol}} = 0.0321 \mathrm{~mol} \]
3. Calculate the moles of $\mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O}$: \[ \text{Moles of } \mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O} = 0.0321 \mathrm{~mol} \]
4. Calculate the mass of $\mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O}$: \[ \text{Molar mass of } \mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O} = 56 + 32 + 7 \times 18 = 277 \mathrm{~g/mol} \] \[ \text{Mass of } \mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O} = 0.0321 \mathrm{~mol} \times 277 \mathrm{~g/mol} = 8.8935 \mathrm{~g} \approx 9 \mathrm{~g} \] Therefore, the correct answer is (9).