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A student wants to reclaim the iron from an 18.0-gram sample of iron(III) oxide, which
occurs according to the reaction shown below:
2Fe2O3(s) — 4Fe(s) + 30,(g)
How many grams of iron can be reclaimed?

Respuesta :

Eduard22sly

Answer:

12.6g of Fe.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2Fe2O3(s) —> 4Fe(s) + 3O2(g)

Next, we shall determine the mass of Fe2O3 that decomposed and the mass of Fe that is produced from the balanced equation.

This is illustrated below:

Mola mass of Fe2O3 = (56 x 2) + (16x3) = 160g/mol

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

Molar mass of Fe = 56g/mol

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

From the balanced equation above,

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

Finall, we can obtain the mass of the Fe produced from the decomposition of 18g of Fe2O3 as follow:

From the balanced equation above,

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

Therefore 18g of Fe2O3 will decompose to produce = (18x224)/320 = 12.6g of Fe.

Therefore, 12.6g of Fe is obtained from 18g of Fe2O3.

CintiaSalazar

Taking into account the reaction stoichiometry, 12.58 grams of Fe are formed when 18 grams of Fe₂O₃ reacts.

Reaction stoichiometry

In first place, the balanced reaction is:

2 Fe₂O₃  → 4 Fe + 3 O₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Fe₂O₃: 2 moles  
  • Fe: 4 moles
  • O₂: 3 moles

The molar mass of the compounds is:

  • Fe₂O₃: 159.7 g/mole
  • Fe: 55.85 g/mole
  • O₂: 32 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • Fe₂O₃: 2 moles× 159.7 g/mole= 319.4 grams
  • Fe: 4 moles× 55.85 g/mole= 223.4 gram
  • O₂: 3 moles× 32 g/mole= 96 grams

Then the following rule of three can be applied: if by reaction stoichiometry 319.4 grams of Fe₂O₃ form 223.3 grams of Fe, 18 grams of Fe₂O₃ form how much mass of Fe?

[tex]mass of Fe=\frac{18 grams of Fe_{2} O_{3} x223.3 grams of Fe}{319.4 grams of Fe_{2} O_{3}}[/tex]

mass of Fe= 12.58 grams

Then, 12.58 grams of Fe are formed when 18 grams of Fe₂O₃ reacts.

Learn more about reaction stoichiometry:

brainly.com/question/24741074

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