Question

An alkaline battery produces electrical energy according to the following equation.

Zn(s) + 2 MnO2(s) + H2O(l) Zn(OH)2(s) + Mn2O3(s)
(a) Determine the limiting reactant if 37.0 g Zn and 21.5 g MnO2 are used. (Type your answer using the format CH4 for CH4.)


(b) Determine the mass of Zn(OH)2 produced.
g

Answer 1

Part a: limiting reactant MnO₂

Part b: 12.43 g of Zn(OH)₂

Step-by-step 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.