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3 years ago

8

how to solve stoichiometry problem in a synthesis reaction with iron metal and oxygen gas, what mass of iron metsl is required t

o produce 456.0g of ferric oxide?

1 answer:

sesenic [268]3 years ago

8 0

Answer:

319.2 g of iron metal

Explanation:

We'll begin by writing the balanced equation for the reaction between iron (Fe) and oxygen (O₂) to produce ferric oxide (Fe₂O₃). This is illustrated below:

4Fe + 3O₂ —> 2Fe₂O₃

Next, we shall determine the mass of Fe that reacted and the mass of Fe₂O₃ produced from the balanced equation. This can be obtained as follow:

Molar mass of Fe = 56 g/mol

Mass of Fe from the balanced equation = 4 × 56 = 224 g

Molar mass of Fe₂O₃ = (56×2) + (16×3)

= 112 + 48

= 160 g/mol

Mass of Fe₂O₃ from the balanced equation = 2 × 160 = 320 g

SUMMARY:

From the balanced equation above,

224 g of Fe reacted to produce 320 g of Fe₂O₃.

Finally, we shall determine the mass of iron metal, Fe, required to produce 456 g of ferric oxide, Fe₂O₃. This can be obtained as follow:

From the balanced equation above,

224 g of Fe reacted to produce 320 g of Fe₂O₃.

Therefore, Xg of Fe will react to produce 456 g of Fe₂O₃ i.e

Xg of Fe = (224 × 456)/320

Xg of Fe = 319.2 g

Thus, 319.2 g of iron metal, Fe is required to produce 456 g of ferric oxide, Fe₂O₃.

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From the question we are told that

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