a) PTHH: HCl + KOH -> KCl + H2O
nHCl= 1/36,5(mol)
nKOH= 1/56 (mol)
Vì: 1/36,5 :1 > 1/56 :1
=> HCl dư, KOH hết, tính theo nKOH
=> nHCl (dư)= 1/36,5 - 1/56\(\approx\) 0,01 (mol)
=> mHCl(dư)\(\approx\) 0,01.36,5 \(\approx\) 0,365(g)
b) nKCl= nKOH= 1/56 (mol)
=> m(muối)=mKCl= 1/56 . 74,5\(\approx\) 1,33(g)