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An alkaline battery produces electrical energy according to this balanced equation:
Zn + 2MnO2 + H2O -> Zn(OH)2 + Mn2O3
Determine the limiting reactant if 25.0 g of Zn and 30.0 g of MnO2 are used. In addition, what mass of Zn(OH)2 will be produced?

A) Zn; 38.0 g Zn(OH)2
B) Zn: 17.2 g Zn(OH)2
C) MnO2; 38.0g Zn(OH)2
D) MnO2; 17.2 g Zn(OH)2

Respuesta :

CintiaSalazar

Considering the reaction stoichiometry and the definition of limiting reagent, the correct answer is option D. MnOâ‚‚; 17.2 grams Zn(OH)â‚‚.

The balanced reaction is:

Zn + 2 MnO₂ + H₂O → Zn(OH)₂ + Mn₂O₃

Reaction stoichiometry

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:  

  • Zn: 1 mole
  • MnOâ‚‚: 2 moles
  • Hâ‚‚O: 1 mole
  • Zn(OH)â‚‚: 1 mole
  • Mnâ‚‚O₃: 1 mole

The molar mass of the compounds present in the reaction is:

  • Zn: 65.37 g/mole
  • MnOâ‚‚: 86.94 g/mole
  • Hâ‚‚O: 18 g/mole
  • Zn(OH)â‚‚: 99.37 g/mole
  • Mnâ‚‚O₃: 157.88 g/mole

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

  • Zn: 1 mole× 65.37 g/mole= 65.37 grams
  • MnOâ‚‚: 2 moles× 86.94 g/mole=  173.88 grams
  • Hâ‚‚O: 1 mole× 18 g/mole= 18 grams
  • Zn(OH)â‚‚: 1 mole× 99.37 g/mole= 99.37 grams
  • Mnâ‚‚O₃: 1 mole× 157.88 g/mole= 157.88 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.

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 65.37 grams of Zn reacts with 173.88 grams of MnOâ‚‚, 25 grams of Zn reacts with how much mass of MnOâ‚‚?

[tex]mass of MnO_{2} =\frac{25 grams of Znx173.88 grams of MnO_{2}}{65.37 grams of Zn}[/tex]

mass of MnOâ‚‚ = 66.498 grams

But 66.498 grams of MnOâ‚‚ are not available, 30 grams are available. Since you have less mass than you need to react with 25 grams of Zn, MnOâ‚‚ will be the limiting reagent.

Mass of Zn(OH)â‚‚ produced

Then, it is possible to determine the Mass of Zn(OH)â‚‚ produced by another rule of three, considering the limiting reactant: if by stoichiometry 173.88 grams of MnOâ‚‚ produce 99.37 grams of Zn(OH)â‚‚, if 30 grams of MnOâ‚‚ react how much mass of Zn(OH)â‚‚ will be formed?

[tex]mass of Zn(OH)_{2}=\frac{30 grams of MnO_{2}x99.37 grams of Zn(OH)_{2} }{173.88 grams of MnO_{2}}[/tex]

mass of Zn(OH)₂= 17.14 grams≅ 17.2 grams

In summary, the mass of Zn(OH)â‚‚ that is produced when 25 grams of Zn react with 30 grams of MnOâ‚‚ is 17.2 grams.

Summary

In summary, the correct answer is option D. MnOâ‚‚; 17.2 grams Zn(OH)â‚‚.

Learn more about reaction stoichiometry:

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darlaengler7

Answer:

this is a long one for a full answer...

the answer is D) MnO2; 17.2 g Zn(OH)2

Explanation:

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