Determine the limiting reactant if 17.5 g Zn and 27.5 g MnO2 are used. Determine the mass of Zn(OH)2 produced.

Question

 I really need help finding the mass. So I included the previous masses.

Answer 1

Balance the equation first, then calculate the g/mol of each molecule. Then try and see if you can figure the rest out. Give it a shot, and then come back with a question and your attempt at solving it if you're still confused

Answer 2

This is NOT an easy question to answer, so I will assist you.  The reaction referred to in the question is a redox reaction that takes place in an alkaline battery. I believe the product of the reaction is ZnO, not Zn(OH)2, which is insoluble. Perhaps your teacher is using the first half reaction below and assuming that ZnO + H2O -->Zn(OH)2. The half reactions are:

 

Zn(s) + 2OH−(aq) ==> ZnO(s) + H2O(l) + 2e− 
2MnO2(s) + H2O(l) + 2e− ==> Mn2O3(s) + 2OH−(aq) 
The overal reaction is:
Zn(s) + 2MnO2(s) <==> ZnO(s) + Mn2O3(s)  This is the balanced equation.  Now for calculation of limiting reactant and mass of ZnO.  (If you need Zn(OH)2, you'll have to re-submit after being sure that's what is needed).

moles Zn = 17.5 g x 1 mole/ 65.4g = 0.268 moles

moles MnO2 = 27.5 g x 1 mole/86.9 g = 0.316 moles

Limiting reactant = MnO2 since stoichiometric mole ratio is 2 moles MnO2 :1 mole Zn.  MnO2 will run out first.

Mass of product produced (either ZnO or Zn(OH)2) will depend on moles MnO2 only.

2 moles MnO2 produces 1 mole of ZnO as determined from the overal reaction.

moles ZnO produced = 0.316 moles MnO2 x 1 mole ZnO/2 moles MnO2 = 0.158 moles ZnO produced

mass ZnO produced = 0.158 moles x 81.4 g/mole = 12.9 grams

 

NOTE:  I think you can follow this approach using Zn(OH)2 as the product, but I'm not sure how that product is arrived at.  Unless, as stated in the beginning the ZnO reacts with water to produce Zn(OH)2.  Most oxides do react with water to produce the metal hydroxide, so you may want to rework the problem using that approach.