Đáp án:

%Fe =56,4%

%Zn = 43,6%

C% HCl =14,6%

Giải thích các bước giải:

 nH2=5,6/22,4=0,25 mol

Gọi nFe=x mol; nZn=y mol

Ta có: 56x + 65y =14,9 g

Pt: Fe + 2HCl -> FeCl2 + H2

      x                                    x  mol

Pt: Zn + 2HCl -> ZnCl2 + H2 

   y                                      y   mol

=> Ta có hệ pt: 56x + 65y =14,9        

                             x + y =0,25 

=> x=0,15; y=0,1 mol

%Fe = 0,15.56.100%/14,9=56,4%

%Zn = 0,1.65.100%/14,9=43,6%

mddHCl = 250.0,5=125g

mHCl = 2(x+y).36,5=2.0,25.36,5=18,25g

C% HCl = 18,25.100%/125=14,6%

Đáp án:

$\%m_{Fe}=56,38\%\\ \%m_{Zn}=46,32\%\\ C\%_{dd\ FeCl_2}=13,67\%\\ C\%_{dd\ ZnCl_2}=9,76\%$

$C\%_{dd\ HCl}=14,6\%$

Giải thích các bước giải:

$n_{H_2}=\dfrac{5,6}{22,4}=0,25(mol)\\ m_{dd\ HCl}=250.0,5=125(g)\\ Đặt\ \begin{cases}n_{Fe}=x(mol)\\n_{Zn}=y(mol)\end{cases}\\ \Rightarrow 56x+65y=14,9\ (1)\\ Fe+2HCl\to FeCl_2+H_2\\ Zn+2HCl\to ZnCl_2+H_2\\ n_{H_2}=n_{Fe}+n_{Zn}=x+y(mol)\\ \Rightarrow x+y=0,25\ (2)\\\ \xrightarrow{\text{Từ (1),(2)}}\begin{cases}x=0,15\\y=0,1\end{cases}\\ \%m_{Fe}=\dfrac{0,15.56}{14,9}.100\%=56,38\%\\ \%m_{Zn}=\dfrac{0,1.65}{14,9}.100\%=43,62\%\\ n_{FeCl_2}=n_{Fe}=0,15(mol)\\ m_{FeCl_2}=0,15.127=19,05(g)\\ n_{ZnCl_2}=n_{Zn}=0,1(mol)\\ m_{ZnCl_2}=0,1.136=13,6(g)\\ m_{dd\ sau}=m_{\text{kim loại}}+m_{dd\ HCl}-m_{H_2}\\ =14,9+125-0,25.2\\ =139,4(g)\\ C\%_{dd\ FeCl_2}=\dfrac{19,05}{139,4}.100\%=13,67\%\\ C\%_{dd\ ZnCl_2}=\dfrac{13,6}{139,4}.100\%=9,76\%$

$n_{HCl}=2.n_{H_2}=0,5(mol)\\ C\%_{dd\ HCl}=\dfrac{0,5.36,5}{125}.100\%=14,6\%$