Answer:
Part a: limiting reactant MnO₂
Part b: 12.43 g of Zn(OH)₂
Explanation:
Part a
Data given:
mass of Zn = 37.0 g
mass of MnO₂ = 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 MnO₂ = 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) 2 mol (87 g/mol)
65 g 174 g
So its clear from the reaction that 65 g Zn react with 174 g of MnO₂.
now if we look at the given amounts the amount MnO₂ is less then the amount of Zn but in actual calculation amount of MnO₂ is more then amount of zinc.
So, for MnO₂ 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 MnO₂ will be consumed completely an it will be limiting reactant.
____________
part b
Data given:
mass of Zn = 37.0 g
mass of MnO₂ = 21.5 g
Mass of Zn(OH)₂ 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 MnO₂ is limiting reactant, so the amount of Zn(OH)₂ will depend on the amount of MnO₂.
So first we convert mass of MnO₂ to moles
Formula Used
no. of moles = mass in grams / molar mass
Mass of MnO₂ = 55 + 2 (16)
Mass of MnO₂ = 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
So its clear from the reaction that 2 mole of MnO₂ gives 1 mole of Zn(OH)₂
then how many moles of Zn(OH)₂ will be produce by 0.25 moles of MnO₂
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
mass in grams = 12.43 g
So,
12.43 g of Zn(OH)₂ will be produce.
