What is the percentage composition of iron in ferric sulphate, given mass of iron is 56 amu?

14%

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28%

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42%

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56%

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Solution

The correct option is B
28%

Chemical formula of Ferric Sulphate - $F e_{2} (S O_{4})_{3}$

The atomic of Fe = 56, S = 32 and O = 16

Molar Mass of the compound $= 56 \times 2 + (16 \times 4 + 32) \times 3 = 400$

Percentage composition of Fe
= $\frac{56 \times 2}{400} \times 100$ = 28%

Suggest Corrections

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