Làm ơn giúp mình ạ!!:r2:r2:r2
1)
mNaOH=200.20%=40(g)
=>nNaOH=40/40=1(mol)
pt: NaOH+HCl--->NaCl+H2O
1______________1(mol)
nHCl=100/36,5~2,74(mol)
Theo pt: nHCl=nNaOH=1(mol)
=>HCl dư
Ta có: mNaCl=1.(23+35,5)=58,5(g)
mHCl dư=100-1.36,5=63,5(mol)
m(dung dịch)=mNaOH+mHCl=100+200=300(g)
=>C%NaCl
=>C%HCl
2)
Gọi nFeCl2 là x
=>nNaOH=2x
nNaCl=2x
Ta có :
C%(NaOH)=2x.40/mddNaOH.100%=20%
=>80x/mddNaOH.100%=20%
=>mddNaOH=400x
C%(feCl2)=127.x/mddFeCl2.100%=10%
=>mddFeCl2=1270x
=>C%NaCl=2x.58,5/(mddNaOH+mddFeCl2-mFe(OH)2).100%
<=>117x/(400x+1270x-x.90).100%~7.4%
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