Question

the mass of oxygen required for the rusting of 4.2g of ironis

Answer 1

Answer:

The formula for rust is Fe203

Therefore, oxygen will be required 6.3g

Explanation:

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Answer 2

Your answer:

The formation of rust:

4Fe + 3O₂ → 2Fe₂O₃

So, 4 moles of iron react with 3 moles of oxygen to give 2 moles of rust

No. of moles of iron in 4.2 g of iron = $\frac{4.2}{56}$ = 0.075 moles of iron

0.075 moles of iron reacts with 0.05625 moles of oxygen to give 0.0375 moles of rust

Mass of 0.05625 moles of oxygen = 0.05625 × 32 = 1.8 g of oxygen is required for rusting of 4.2 g of iron

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