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An iron nail weighing 40.0 g is placed in 200.0 g of copper(II) sulfate, which reacts to form pure copper and iron(III) sulfate by the following reaction:
__Fe(s) + __CuSO4(s) → __Fe2(SO4)3(s) + __Cu(s)
a. Fill in the missing coefficients to balance the equation.
b. What is the limiting reagent and the mass of iron(III) sulfate that is formed?
c. After the reaction goes to completion, what is the mass of the reactant that remains?

Respuesta :

superaustin5660

Answer:

Fe(s) + CuSO4 (aq) ! FeSO4 (aq) + Cu(s)

Fe(s) + Cu+2(aq) ! Fe+2(aq) + Cu(s)

(2) 2Fe(s) + 3CuSO4 (aq) ! Fe2(SO4)3 (aq) + 3Cu(s)

2Fe(s) + 3Cu+2(aq) ! 2Fe+3(aq) + 3Cu(s)

Explanation:

CintiaSalazar

Taking into account the reaction stoichiometry:

a. The balanced reaction is 2 Fe + 3 CuSO₄ → Fe₂(SO₄)₃ + 3 Cu

b. Fe will be the limiting reagent.

c. After the reaction goes to completion, the mass of the reactant that remains is 28.61 grams.

Reaction stoichiometry

In first place, the balanced reaction is:

2 Fe + 3 CuSO₄ → Fe₂(SO₄)₃ + 3 Cu

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:

  • Fe: 2 moles
  • CuSO₄: 3 moles
  • Fe₂(SO₄)₃: 1 mole
  • Cu: 3 moles

The molar mass of the compounds is:

  • Fe: 55.85 g/mole
  • CuSO₄: 159.54 g/mole
  • Fe₂(SO₄)₃: 399.7 g/mole
  • Cu: 63.54 g/mole

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

  • Fe: 2 moles ×55.85 g/mole= 111.7 grams
  • CuSO₄: 3 moles ×159.54 g/mole= 478.62 grams
  • Fe₂(SO₄)₃: 1 mole ×399.7 g/mole= 399.7 grams
  • Cu: 3 moles ×63.54 g/mole= 190.62 grams

Limiting reagent

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

Limiting reagent in this case

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 478.62 grams of CuSO₄ reacts with 111.7 grams of Fe, 200 grams of CuSO₄ reacts with how many mass of Fe?

[tex]mass of Fe=\frac{200 grams of CuSO_{4} x111.7 grams of Fe}{478.62 grams of CuSO_{4}}[/tex]

mass of Fe= 46.67 grams

But 46.67 grams of Fe are not available, 40 grams are available. Since you have less mass than you need to react with 200 grams of CuSO₄, Fe will be the limiting reagent.

Mass of the reactant that remains

First you calculate the amount of excess reactant that reacts with the entire limiting amount of reactant.

So it is possible to use a simple rule of three as follows: if by stoichiometry 111.7 grams of Fe reacts with 478.62 grams of CuSO₄, 40 grams of Fe reacts with how many mass of CuSO₄?

[tex]mass of CuSO_{4}=\frac{478.62 of CuSO_{4} x40 grams of Fe}{111.7 grams of Fe}[/tex]

mass of CuSO₄= 171.39 grams

You have 200 grams of CuSO₄. Then, the amount of excess of the compound is calculated as:

200 grams - 171.39 grams= 28.61 grams

After the reaction goes to completion, the mass of the reactant that remains is 28.61 grams.

Learn more about the reaction stoichiometry and limiting reagent:

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