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