Question

How many grams of Fe3O4 are required to react completely with 300 grams of H2? Fe3O4 + H2=Fe + H2O

Answer 1

Answer:

8596 g of $Fe_3O_4$ is present

Explanation:

Given the equation is

$\mathrm{Fe}_{3} \mathrm{O}_{4}+\mathrm{H}_{2}=\mathrm{Fe}+\mathrm{H}_{2} \mathrm{O}$

On balancing the equation, we get,

$\mathrm{Fe}_{3} \mathrm{O}_{4}+4 \mathrm{H}_{2}=3 \mathrm{Fe}+4 \mathrm{H}_{2} \mathrm{O}$

So, 1 mole of $Fe_3O_4$ reacts with 4 moles of $H_2$

Therefore the ratio is 1:4

Moles of $H_2$used in the reaction = 300/0.02 = 148.5 moles

$Fe_3O_4$present in the reaction =

$148.5 \times \frac{1}{4} \times 231.55=8596 g$

Therefore, 8596 g of $Fe_3O_4$ is present

Answer 2

The mass of Fe₃O₄ required to react completely with 300 g of H₂ is 8700 g

Balanced equation

Fe₃O₄ + 4H₂ —> 3Fe + 4H₂O

Molar mass of Fe₃O₄ = (3×56) + (4×16) = 232 g/mol

Mass of Fe₃O₄ from the balanced = 1 × 232 = 232 g

Molar mass of H₂ = 2 × 1 = 2 g/mol

Mass of H₂ from the balanced equation = 4 × 2 = 8 g

SUMMARY

From the balanced equation above,

8 g of H₂ reacted with 232 g of Fe₃O₄

How to determine the mass of Fe₃O₄ needed

From the balanced equation above,

8 g of H₂ reacted with 232 g of Fe₃O₄

Therefore,

300 g of H₂ will react with = (300 × 232) / 8 = 8700 g of Fe₃O₄

Thus, 8700 g of Fe₃O₄ is needed for the reaction.

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