Đáp án:
$a/$
$Fe+2HCl \to FeCl_2+H_2$
$Mg+2HCl \to MgCl_2+H_2$
$2NaOH+FeCl_2 \to 2NaCl+Fe(OH)_2$
$2NaOH+MgCl_2 \to 2NaCl+Mg(OH)_2$
$b/C\%FeCl_2=15,76\%$
$c/C\%NaCl=10,36\%$
giải thích các bước giải:
Gỉa sử có 1mol $HCl $ phản ứng
$mHCl=1.36,5=36,5g$
$mddHCl =\frac{36,5}{20\%}=182,5g$
Gọi số mol $Fe$: x
$Mg$: y
$Fe+2HCl \to FeCl_2+H_2$
$Mg+2HCl \to MgCl_2+H_2$
⇒$2x+2y=1⇔x+y=0,5$
⇒$x=0,5-y$(1)
Bảo toàn nguyên tố "H"
⇒$2nH_2=nHCl⇒nH_2=0,5$
$mX=56x+24y=56.(0,5-y)+24y=-32y+28$
$mddsau=mX+mddHCl-mH_2=-32y+209,5g$
$C\%MgCl_2 =\frac{95y}{-32y+209,5}$
⇔$11,787\%=\frac{95y}{-32y+209,5}$$
⇒$y=0,25$
(1)⇒$X=0,5-0,25=0,25$
$C\%FeCl_2=\frac{0,25.127}{-32.0,25+209,5}.100=15,76\%$
c/
$2NaOH+FeCl_2 \to 2NaCl+Fe(OH)_2$
$2NaOH+MgCl_2 \to 2NaCl+Mg(OH)_2$
$nNaOH=0,25.2+0,25.2=1$
$mddNaOH=\frac{1.40}{10\%}=400g$
$mFe(OH)_2=0,25.90=22,5g$
$mMg(OH)_2=0,25.58=14,5g$
$mddsau=mddNaOH+mddY-mKT=400+(-32.0,25+209,5)-22,5-14,5=564,5g$
$C\%NaCl=\frac{1.58,5}{564,5}.100=10,36\%$
