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1) Chemical reaction:

Fe2O3 + 2Al ---> Al2O3 + 2Fe

2) molar ratios

1 mol Fe2O3 : 2Al : 1 mol Al2O3 : 2 mol Fe

3) Convert 15.0 g of iron into moles

atomic mass Fe = 55.8 g/mol

moles = mass in grams / atomic mass = 15.0 g / 55.8 g/mol = 0.269 mol

4) Use proportions to determine the moles of Fe2O3, Al, and Al2O3

a) 1mol Fe2O3 / 2 mol Fe = x / 0.269 mol Fe

x =

=> x = 0.269 mol Fe * 1 mol Fe2O3 / 2 mol Fe = 0.134 mol Fe2O3

b) 2 mol Al / 2 mol Fe = x / 0.269 mol Fe

=> x = 0.269 mol Al

c) 2 mol Fe / 1 mol Al2O3 = 0.269 mol Fe / x

=> x = 0.269 mol Fe * 1 mol Al2O3 / 2 mol Fe

x = 0.134 mol Al2O3

5) Convert moles to grams

a) Fe2O3

molar mass Fe2O3 = 2* 55.8 g/mol + 3*16g/mol = 159.6 g/mol

mass = molar mass * number of moles

mass = 159.6 g/mol * 0.134 mol = 21.4 g

b) Al

atomic mass = 27.0 g/mol

mass = number of moles * atomic mass = 0.269 mol * 27.0 g/mol = 7.26 g

c) Al2O3

molar mass = 2 * 27.0 g/mol + 3*16.0 g/mol = 102.0g/mol

mass Al2O3 = numer of moles * molar mass = 0.134 mol * 102.0 g/mol = 13.7 g

Answers:

21.4 g Fe2O3

7.26 g Al

13.7 g Al2O3
CintiaSalazar

Taking into account the reaction stoichiometry:

  • 21.44 grams of Fe₂O₃ and 6.18 grams of Al is required to produce 15 grams of Fe.
  • 12.62 grams of Al₂O₃ is formed.

Reaction stoichiometry

In first place, the balanced reaction is:

Fe₂O₃ + 2 Al → Al₂O₃ + 2 Fe

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: 1 mole
  • Al: 2 moles
  • Al₂O₃: 1 mole
  • Fe: 2 moles

The molar mass of the compounds is:

  • Fe₂O₃: 159.7 g/mole
  • Al: 23 g/mole
  • Al₂O₃: 94 g/mole
  • Fe: 55.85 g/mole

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

  • Fe₂O₃: 1 mole ×159.7 g/mole= 159.7 grams
  • Al: 2 moles ×23 g/mole= 46 grams
  • Al₂O₃: 1 mole ×94 g/mole= 94 grams
  • Fe: 2 moles ×55.85 g/mole= 111.7 grams

Mass of Fe₂O₃ required

The following rule of three can be applied: if by reaction stoichiometry 111.7 g of Fe are produced from 159.7 g of Fe₂O₃, 15 g of Fe are produced from how much mass of Fe₂O₃?

[tex]mass of Fe_{2} O_{3} =\frac{15 grams of Fex159.7 gramsof Fe_{2} O_{3}}{111.7 grams of Fe}[/tex]

mass of Fe₂O₃= 21.44 grams

Finally, 21.44 grams of Fe₂O₃ is required to produce 15 grams of Fe.

Mass of Al required

The following rule of three can be applied: if by reaction stoichiometry 111.7 g of Fe are produced from 46 g of Al, 15 g of Fe are produced from how much mass of Al?

[tex]mass of Al =\frac{15 grams of Fex46 gramsof Al}{111.7 grams of Fe}[/tex]

mass of Al= 6.18 grams

Finally, 6.18 grams of Al is required to produce 15 grams of Fe.

Mass of Al₂O₃ formed

The following rule of three can be applied: If by stoichiometry of the reaction 111.7 grams of Fe are produced together with 94 grams of Al₂O₃, 15 grams of Fe are produced together with how much mass of Al₂O₃?

[tex]mass of Al_{2} O_{3} =\frac{15 grams of Fex94 grams of Al_{2} O_{3}}{111.7 grams of Fe}[/tex]

mass of Al₂O₃= 12.62 grams

Then, 12.62 grams of Al₂O₃ is formed.

Learn more about the reaction stoichiometry:

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