a. PTHH:

\(Ag+HCl--\times-->\)

\(Zn+2HCl--->ZnCl_2+H_2\)

b. Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\) (phần này mình sửa lại phần số mol)

Theo PT: \(n_{Zn}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{Zn}=0,1.65=6,5\left(g\right)\)

\(\Rightarrow m_{Ag}=25,8-6,5=19,3\left(g\right)\)

c. \(\%_{m_{Zn}}=\dfrac{6,5}{25,8}.100\%=25,19\%\)

\(\%_{m_{Ag}}=100\%-25,19\%=74,81\%\)

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1=n_{Zn}\\ n_{ZnO}=\dfrac{14,6-6,5}{81}=0,1mol\\ C\%=\dfrac{0,2\cdot136}{175,6+14,6-0,2}=14,32\%\)

PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)=n_{Fe}\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,2\left(mol\right)\\\%m_{Fe}=\dfrac{0,1\cdot56}{12}\cdot100\%\approx46,67\%\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{HCl}}=\dfrac{0,2}{0,2}=1\left(M\right)\\\%m_{Cu}=53,33\%\end{matrix}\right.\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

Cu không phản ứng

\(nH_2=nFe=\dfrac{2,24}{22,4}=0,1mol\)

\(\rightarrow mFe=0,1.56=5,6gam\)

\(\rightarrow\%mFe=\dfrac{5,6}{12}.100\%=46,\left(6\right)\%\)

\(\rightarrow\%mCu=100\%-46,\left(6\right)\%=53,\left(3\right)\%\)

c)

\(CM_{HCl}=\dfrac{0,1.2}{0,2}=1M\)

a Fe + 2HCl ---> FeCl2 + H2

nH2= 2,24 /22,4=0,1 mol

nFe=nH2=0,1 mol

mFe=0,1.56=5.6 g

mCu=12-5,6=6,4g

%mfe=5,6/12 .100%=46,67

%mCu=100%-46,67%=53,33%

n HCl=0,05 mol

CM ddHCl= 0,2/0,2=1M

Fe+ H2SO4 -> FeSO4 + H2

0,1___0,1______0,1__0,1(mol)

FeO+ H2SO4-> FeSO4 + H2O

=>mFe= 0,1.56=5,6(g)

=>%mFe=(5,6/12.8).100=43.75%

=>%mFeO=56,25%

\(a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ ZnO+2HCl\to ZnCl_2+H_2O\\ b,n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ \Rightarrow n_{Zn}=0,15(mol)\Rightarrow m_{Zn}=0,15.65=9,75(g)\\ \Rightarrow \%_{Zn}=\dfrac{9,75}{25,95}.100\%=37,57\%\\ \Rightarrow \%_{ZnO}=(100-37,57)\%=62,43\%\\ c,n_{ZnO}=\dfrac{25,95-9,75}{81}=0,2(mol)\\ \Rightarrow n_{HCl}=2.0,15+2.0,2=0,7(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{0,7.36,5}{12\%}=212,92(g)\)

Chào em, em đăng bài nhầm môn, em đăng sang môn Hóa nha!

a.\(Fe+S\rightarrow\left(t^o\right)FeS\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(FeS+2HCl\rightarrow FeCl_2+H_2S\)

b.\(n_{hhk}=\dfrac{4,48}{22,4}=0,2mol\)

\(Fe+S\rightarrow\left(t^o\right)FeS\)

Ta thu được hh khí --> S hết, Fe dư

Gọi \(\left\{{}\begin{matrix}n_{Fe}=x\\n_S=y\end{matrix}\right.\)

\(\rightarrow n_{FeS}=n_{Fe}=n_S\rightarrow n_{Fe\left(dư\right)}=x-y\) ( mol )

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(x-y\)                            \(x-y\)        ( mol )

\(FeS+2HCl\rightarrow FeCl_2+H_2S\)

  y                                          y       ( mol )

Ta có: \(\left(x-y\right)+y=0,2\)

           \(\Leftrightarrow x=0,2\)

Ta có:\(56x+32y=14,4\)

        \(\Leftrightarrow56.0,2+32y=14,4\)

        \(\Leftrightarrow y=0,1\)

\(\rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,2.56}{14,4}.100=77,77\%\\\%m_S=100\%-77,77\%=22,23\%\end{matrix}\right.\)

nBr2 = 32/160 = 0,2 (mol)

PTHH: C2H4 + Br2 -> C2H4Br2

Mol: 0,2 <--- 0,2

VC2H4 = 0,2 . 22,4 = 4,48 (l)

%VC2H4 = 4,48/6,2 = 72,25%

%VCH4 = 100% - 72,25% = 27,75%

a)

$Zn + H_2SO_4 \to ZnSO_4 + H_2$
$Mg + H_2SO_4 \to MgSO_4 + H_2$

b)

Gọi $n_{Zn} = a(mol) ; n_{Mg} = b(mol) \Rightarrow 65a + 24b = 19,85(1)$

Theo PTHH : 

$n_{H_2} = a + b = \dfrac{8,96}{22,4} = 0,4(2)$

Từ (1)(2) suy ra : a = 0,25 ; b = 0,15

$\%m_{Zn} = \dfrac{0,25.65}{19,85}.100\% = 81,9\%$
$\%m_{Mg} = 100\% - 81,9\% = 18,1\%$

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ a,PTHH:Fe+2HCl\to FeCl_2+H_2\\ b,n_{Fe}=n_{H_2}=0,1(mol)\\ \Rightarrow m_{Fe}=0,1.56=5,6(g)\\ \Rightarrow \%_{Fe}=\dfrac{5,6}{12}.100\%=46,67\%\\ \Rightarrow \%_{Cu}=100\%-46,67\%=53,33\%\\ c,n_{HCl}=2n_{H_2}=0,2(mol)\\ \Rightarrow C_{M_{HCl}}=\dfrac{0,2}{0,2}=1M\)