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 17.5 g Zn and 31.0 g MnO2 are used.

(Type your answer using the format CH4 for CH4.)

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

a) Given reaction:

Zn + 2MnO2 + H2O → Zn (OH) 2 + Mn2O3

1 mole of Zn combines with 2 moles of MnO2

Now:

# moles Zn present = 17.5/65.38 = 0.2677 moles

# moles of MnO2 present = 31.0/86.94 = 0.3566 moles

Since # moles of MnO2 is less than Zn i. e. it is not present in the 2:1 ratio (MnO2:Zn), MnO2 will be the limiting reagent

b) Based on stoichiometry:

2 moles of MnO2 produces 1 mole of Zn (OH) 2

Thus, moles of Zn (OH) 2 produced form the limiting reactant = 0.3566/2 = 0.1783 moles

Mass of Zn (OH) 2 = 0.1783*99.42 = 17.73 g