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/

$2NaOH+FeCl_2 \to 2NaCl+Fe(OH)_2$

$2NaOH+MgCl_2 \to 2NaCl+Mg(OH)_2$

$nNaOH=0,25.2+0,25.2=1$

$mddNaOH=\frac{1.40}{10\%}=400g$

$mFe(OH)_2=0,25.90=22,5g$

$mMg(OH)_2=0,25.58=14,5g$

$mddsau=mddNaOH+mddY-mKT=400+(-32.0,25+209,5)-22,5-14,5=564,5g$

$C\%NaCl=\frac{1.58,5}{564,5}.100=10,36\%$