nK=0,25(mol)
2K+2H2O→2KOH+H2↑
Theo PT nKOH=nK=0,25(mol)
=> mKOH=0,25.56=14(g)
Theo PT nH2=0,5nK=0,5.0,25=0,125(mol)
=> mH2=0,125.2=0,25(g)
mdd=9,75+400−0,25=409,5(g)
mNaOH=400.15%=60(g)
C%KOH=\(\dfrac{14}{409,5}.100\)=3,419%
C%NaOH=\(\dfrac{60}{409,5}.100\)=14,652%