giải:
nNa= 4.6/23 = 0.2 mol
nK = 3.9/39 = 0.1 mol
Na + H2O --> NaOH + 1/2H2
nH2= 0.1+0.05 = 0.15 mol
VH2= 0.15*22.4 = 3.36l
mNaOH = 0.2*40 = 8g
mKOH = 0.1*56 = 5.6 g
mdungdich sau phản ứng = mhh + mH2O - mH2 = 4.6 + 3.9 + 91.5-0.15*2=99.7g
C%NaOH = 8/99.7*100% = 8.02%
C%KOH = 5.6/99.7*100% = 5.67%