mH2O= 100.1=100(g)
PTHH: K + H2O -> KOH + 1/2 H2
nK= 0,1(mol)
nH2[tex]\approx 5,556 (mol)[/tex]
=> 0,1/ 1< 5,556/1 => K hết, H2O dư, tính theo nK
=> nKOH= 0,1(mol) => mKOH= 5,6(g)
mddKOH= mK + mH2O - mH2 = 3,9+100- 0,05.2= 103,8(g)
=> C%ddKOH= (5,6/103,8).100[tex]\approx[/tex] 5,395%