If i perform this reaction by combining 25.0 grams of lioh with an excess of fe(no3)3, how much fe(oh)3 will i be able to make

3LiOH+Fe(NO3)3--›Fe(OH)3↓+3LiNO3
n(LiOH)=m/M=(25g)/(24g/mol)=1,0417mol
n(Fe(OH)3)=1/3n(LiOH)=0,3472mol
m(Fe(OH)3)=n*M=0,3472mol*107g/mol=37,15g

Alpha particles have positive charge, beta particles have negative charge, and answer 4 just doesn’t make sense. thus, you are left with only one possible answer.