CÂU 1

Giải thích các bước giải:

 a)

Fe+2HCl->FeCl2+H2

nH2=3,36/22,4=0,15 mol

mFe=0,15x56=8,4 g

mAg=15-8,4=6,6 g

b)

%mFe=8,4/15x100%=56%

%mAg=100-56=44%

c)

nHCl=2nH2=0,3 mol

m dd HCl=0,3x36,5/15,6%=70,19 g

CÂU 2

 a)

Fe+2HCl->FeCl2+H2

Zn+2HCl->ZnCl2+H2

nH2=11,2/22,4=0,5 mol

gọi a là số mol Fe b là số mol Zn

56a+65b=29,8

a+b=0,5

=>a=0,3 b=0,2

%mFe=0,3x56/29,8x100%=56,38%

%mZn=100-56,38=43,62%

b)

nHCl=0,5x2=1 mol

CMHCl=1/0,6=5/3 M

Em tham khảo nha :

\(\begin{array}{l}
1)\\
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,15mol\\
{m_{Fe}} = 0,15 \times 56 = 8,4g\\
{m_{Ag}} = 15 - 8,4 = 6,6g\\
b)\\
\% Fe = \dfrac{{8,4}}{{15}} \times 100\%  = 56\% \\
\% Ag = 100 - 56 = 44\% \\
c)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,3mol\\
{m_{HCl}} = 0,3 \times 36,5 = 10,95g\\
{m_{{\rm{dd}}HCl}} = \dfrac{{10,95 \times 100}}{{15,6}} = 70,2g\\
2)\\
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{11,2}}{{22,4}} = 0,5mol\\
hh:Fe(a\,mol),Zn(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,5\\
56a + 65b = 29,8
\end{array} \right.\\
 \Rightarrow a = 0,3;b = 0,2\\
{m_{Fe}} = 0,3 \times 56 = 16,8g\\
\% Fe = \dfrac{{16,8}}{{29,8}} \times 100\%  = 56,4\% \\
\% Zn = 100 - 56,4 = 43,6\% \\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 1mol\\
{C_{{M_{HCl}}}} = \dfrac{1}{{0,6}} = \dfrac{5}{3}M
\end{array}\)