Answer:
Part a: limiting reactant MnOâ‚‚
Part b: 12.43 g of Zn(OH)â‚‚
Explanation:
Part a
Data given:
mass of Zn = 37.0 g
mass of MnOâ‚‚ = 21.5 g
Limiting reactant = ?
Given Reaction
    Zn(s) + 2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)
To determine the limiting reactant first we will look at the reaction
     Zn(s)  +  2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)
     1 mol   2 mol
Convert moles to mass
molar mass of Zn = 65.4 g/mol
Mass of MnOâ‚‚ = 55 + 2 (16) = 55 + 32 = 87 g/mol
So,
    Zn(s)      +    2 MnO₂(s)   +   H₂O(l) ----------> Zn(OH)₂(s) + Mn₂O₃(s)
1 mol (65.4 g/mol) Â Â 2 mol (87 g/mol)
    65 g             174 g
So its clear from the reaction that 65 g Zn react with 174 g of MnOâ‚‚.
now if we look at the given amounts the amount MnOâ‚‚ is less then the amount of Zn but in actual calculation amount of MnOâ‚‚ is more then amount of zinc.
So, for MnOâ‚‚ if we calculate the needed amount of zinc
So apply unity formula
      65 g Zn react ≅ 174 g of MnO₂
      X g of Zn ≅ 21.5 g of MnO₂
Do cross multiplication
      X g Zn react = 65 g x 21.5 g / 174 g
      X g of Zn ≅ 8.032 g
So, 8.032 g of zinc will react out of 37.0 grams. the remaining will be in excess.
So MnOâ‚‚ will be consumed completely an it will be limiting reactant.
____________
part b
Data given:
mass of Zn = 37.0 g
mass of MnOâ‚‚ = 21.5 g
Mass of Zn(OH)â‚‚ Â produced = ?
Given Reaction
    Zn(s) + 2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)
Solution:
As from the part A we come to know that MnOâ‚‚ is limiting reactant, so the amount of Zn(OH)â‚‚ will depend on the amount of MnOâ‚‚.
So first we convert mass of MnOâ‚‚ to moles
Formula Used
    no. of moles = mass in grams / molar mass
Mass of MnOâ‚‚ = 55 + 2 (16)
Mass of MnOâ‚‚ = 55 + 32 = 87 g/mol
Put values in the above equation
    no. of moles = 21.5 g / 87 g/mol
    no. of moles = 0.25 moles
Now,
Look at the reaction
     Zn(s)  +  2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)
            2 mol                    1 mol Â
So its clear from the reaction that 2 mole of MnOâ‚‚ gives 1 mole of Zn(OH)â‚‚
then how many moles of Zn(OH)â‚‚ will be produce by 0.25 moles of MnOâ‚‚
So.
apply unity formula
     2 mol of MnO₂ ≅ 1 mole of Zn(OH)₂
      0.25 moles of MnO₂  ≅ X mole of Zn(OH)₂
Do cross multiplication
      X mole of Zn(OH)₂ = 1 mole x 0.25 mol x  / 2 mol
     X mole of Zn(OH)₂  ≅ 0.125
Now Conver moles of  Zn(OH)₂  to mass
Formula used
     mass in grams = no. of moles x molar mass
Molar mass of  Zn(OH)₂
Molar mass of  Zn(OH)₂ = 65.4 + 2 (16 + 1)
Molar mass of  Zn(OH)₂ = 65.4 + 2 (17)
Molar mass of  Zn(OH)₂ = 65.4 + 34 = 99.4 g/mol
Put values in above equation
    mass in grams = 0.125 mol x 99.4 g/mol Â
    mass in grams = 12.43 g
So,
12.43 g of Zn(OH)â‚‚ Â will be produce.
