Câu 1: nNaOH=1(mol)
nH2SO4=0,02(mol)
PTHH: 2 NaOH + H2SO4 -> Na2SO4 +2 H2O
Vì: 0,02/1 < 1/2
=> NaOH dư, H2SO4 hết, tish theo nH2SO4
nNa2SO4=nH2SO4=0,02(mol)
nNaOH(dư)=1-0,02.2=0,96(mol)
Vddsau=200+500=700(ml)=0,7(l)
=>CMddNa2SO4= 0,02/0,7=1/35(M)
CMddNaOH(dư)=0,96/0,7= 48/35(M)
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Bài 3: mBaCl2=200.10,4=20,8(g) -> nBaCl2=0,1(mol)
PTHH: BaCl2 + H2SO4 -> BaSO4 + 2 HCl
nH2SO4=nBaSO4=nBaCl2=0,1(mol) ; nHCl=0,2(mol)
mHCl=0,2.36,5=7,3(g)
VddH2SO4= 0,1/ 0,1= 1(l)=1000(ml)
=> mddH2SO4=1000.1,14=1140(g)
=> mddHCl= mddH2SO4 + mddBaCl2- mBaSO4= 200+ 1140 - 0,1.233= 1316,7(g)
C%ddHCl=(7,3/1316,7).100=0,554%
Bài 2: mHCl=3,65%. 200=7,3(g) -> nHCl=7,3/36,5=0,2(mol)
PTHH: Fe +2 HCl -> FeCl2 + H2
0,2______0,4_____0,2____0,2(mol)
m=mFe=0,2.56=11,2(g)
mddsau= mFe+mddHCl-mH2=5,6+200-0,2.2=205,2(g)
mFeCl2=0,2.127=25,4(g)
=>C%ddFeCl2=(25,4/205,2).100=12,378%
