1.
NaOH+HCl→NaCl+H2O
mNaOH=40*30/100=12g
nNaOH=12/40=0,3mol
mHCl=0,3*36,5=10,95g
C%HCl=10,95/100*100%=10,95%
mdd spu=40+100=140g
mNaCl=0,3*58,5=17.55g
C%NaCl=17,55/140*100%=12,54%
2. Fe+2HCl→FeCl2+H2
nFe=11,2/56=0,2mol
nHCl=0,1*2=0,2mol
Theo PT nFe/nHCl=1/2
Theo đề bài nFe/nHCl=1/1
Vậy Fe dư
⇒nH2=1/2nHCl=0,1mol
⇒VH2=0,1*22,4=2,24l
nFe dư=0,2-0,1=0,1mol
mFe dư=0,1*56=5,6g
Giả sử Vdd trước = Vdd spu=0,1l
CMFeCl2=0,1/0,1=1M