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Based on the limiting reactant, the mass of Zn(OH)2 produced is 38.0 g
What is a limiting reactant?
A limiting reactant is used up in a reaction and the reaction stops afterwards.
Equation of reaction:
- Zn + 2MnO2 + H2O → Zn(OH)2 + Mn2O3
From the equation of the reaction, 1 mole of Zn reacts with 2 moles of MnO2 to produce 1 mole of Zn(OH)2
Molar mass of Zn = 65 g
Molar mass of MnO2 = 56 g
Molar mass of Zn(OH)2 = 99.0 g
Moles of An reacting = 25/65 = 0.38 moles
Moles of MnO2 reacting = 30/56 = 0.5
Therefore, MnO2 is the limiting reactant.
Mass of Zn(OH)2 produced = 25 × 99/65
Mass of Zn(OH)2 produced = 38.0 g of Zn(OH)2
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Taking into account the reaction stoichiometry and limiting reagent, 17.14 grams of Zn(OH)₂ will be produced if 25.0g Zn and 30.0g MnO₂ react.
Reaction stoichiometry
In first place, the balanced reaction is:
Zn + 2 MnO₂ + H₂O → Zn(OH)₂ + Mn₂O₃
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
- MnO₂: 2 moles
- H₂O: 1 mole
- Zn(OH)₂: 1 mole
- Mn₂O₃: 1 mole
The molar mass of the compounds is:
- Zn: 65.37 g/mole
- MnO₂: 86.94 g/mole
- H₂O: 18 g/mole
- Zn(OH)₂: 99.37 g/mole
- Mn₂O₃: 157.88 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Zn: 1 mole ×65.37 g/mole= 65.37 grams
- MnO₂: 2 moles ×86.94 g/mole= 173.88 grams
- H₂O: 1 mole ×18 g/mole= 18 grams
- Zn(OH)₂: 1 mole ×99.37 g/mole= 99.37 grams
- Mn₂O₃: 1 mole ×157.88 g/mole= 157.88 grams
Definition of 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 reaction of this reaction
To determine the limiting reagent, it is possible to use the reaction stoichiometry of the reaction and a simple rule of three as follows: if by stoichiometry 65.37 grams of Zn reacts with 173.88 grams of MnO₂, 25 grams of Zn react with how much mass of MnO₂?
[tex]mass of MnO_{2} =\frac{25 grams of Znx173.88 grams of MnO_{2} }{65.37 grams of Zn}[/tex]
mass of MnO₂= 66.5 grams
But 66.5 grams of MnO₂ are not available, 30 grams are available. Since you have less mass than you need to react with 30 grams of Zn, MnO₂ will be the limiting reagent.
Mass of zinc hydroxide Zn(OH)₂ produced
The following rules of three can be applied, considering the limiting reaction: if by reaction stoichiometry 173.88 grams of MnO₂ form 99.37 grams of Zn(OH)₂, 30 grams of MnO₂ form how much mass of Zn(OH)₂?
[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
Finally, 17.14 grams of Zn(OH)₂ will be produced if 25.0g Zn and 30.0g MnO₂ react.
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