a/ PTHH: Mg + 2HCl ===> MgCl₂ + H₂

x 2x x x

Fe + 2HCl ===> FeCl₂ + H₂

y 2y y y

Gọi số mol Mg, Fe lần lượt là x, y

Lập các số mol theo phương trình

n_H2 = 8,96 / 22,4 = 0,4 mol

Theo đề ra, ta có hệ phương trình:

\(\begin{cases}24x+56y=12,8\\x+y=0,4\end{cases}\)=>\(\begin{cases}x=0,3\\y=0,1\end{cases}\)

=> m_Mg = 0,3 x 24 = 7,2 gam

mFe = 0,1 x 56 = 5,6 gam

b/ \(\sum nHCl\) = 0,8 mol

=> V_HCl = 0,8 / 2 = 0,4 lít = 400ml

c/ PTHH: MgCl₂ + 2NaOH ===> Mg(OH)₂ + 2NaCl

0,3 0,6 0,3

FeCl₂ + 2NaOH ===> Fe(OH)₂ + 2NaCl

0,1 0,2 0,1

=> \(\sum m\downarrow\) = 0,3 x ( 24 + 16 x 2 + 2) + 0,1 x ( 56 + 16 x 2 + 2) = 26,4 gam

Đặt \(n_{Mg}=x(mol);n_{Al}=y(mol)\Rightarrow 24x+27y=7,8(1)\)

\(n_{H_2}=\dfrac{8,96}{22,4}=0,4(mol)\\ PTHH:Mg+2HCl\to MgCl_2+H_2\\ 2Al+6HCl\to 2AlCl_3+3H_2\\ \Rightarrow x+1,5y=0,4(2)\\ (1)(2)\Rightarrow x=0,1(mol);y=0,2(mol)\\ a,m_{Mg}=0,1.24=2,4(g)\\ m_{Al}=7,8-2,4=5,4(g)\\ b,\Sigma n_{HCl}=2x+3y=0,8(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,8}{2}=0,4(l)\)

\(c,PTHH:MgCl_2+2NaOH\to Mg(OH)_2\downarrow+2NaCl\\ AlCl_3+3NaOH\to Al(OH)_3\downarrow+3NaCl\\ n_{MgCl_2}=n_{Mg(OH)_2}=x=0,1(mol)\\ n_{Al(OH)_3}=n_{AlCl_3}=y=0,2(mol)\\ \Rightarrow m_{Mg(OH)_2}=0,1.58=5,8(g)\\ m_{Al(OH)_3}=0,2.78=15,6(g)\\ m_{kết tủa}=5,8+15,6=21,4(g)\)

Fe + 2HCl -> FeCl2 + H2

          0.2                     0.1

FeO + 2HCl -> FeCl2 + H2O

 0.1       0.2

a.\(nH2=\dfrac{2.24}{22.4}=0.1mol\)

\(\%mFe=\dfrac{0.1\times56\times100}{12.8}=43.8\%\)

\(\%mFeO=100-43.8=56.2\%\)

b.\(nFeO=\dfrac{12.8-\left(0.1\times56\right)}{56+16}=0.1mol\)

\(V_{HCl}=\dfrac{0.2+0.2}{2}=0.2l\)

a) \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

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

\(Đặt:n_{Zn}=x\left(mol\right);n_{Fe}=y\left(mol\right)\)

\(n_{H_2}=0,4\left(mol\right)\)

Theo đề ta có hệ \(\left\{{}\begin{matrix}65x+56y=24,2\\x+y=0,4\end{matrix}\right.\)

=> x=0,2 ; y=0,2

\(\%m_{Zn}=\dfrac{0,2.65}{24,2}.100=53,72\%;\%m_{Fe}=46,28\%\)

b)Bảo toàn nguyên tố H: \(n_{HCl}=2n_{H_2}=0,8\left(mol\right)\)

=> \(V_{HCl}=\dfrac{0,8}{2,5}=0,32\left(l\right)\)

c) \(n_{FeCl_2}=0,2\left(mol\right);n_{ZnCl_2}=0,2\left(mol\right)\)

=> \(CM_{FeCl_2}=\dfrac{0,2}{0,32}=0,625\left(mol\right)\)

\(CM_{ZnCl_2}=\dfrac{0,2}{0,32}=0,625\left(mol\right)\)

Fe+2HCl->FeCl₂+H₂

x---2x-----------x

Mg+2HCl->MgCl₂+H₂

y------2y-----------y

Ta có :

\(\left\{{}\begin{matrix}56x+24y=24\\x+y=\dfrac{13,44}{22,4}\end{matrix}\right.\)

=>x=0,3 mol, y=0,3 mol

=>%m Fe=\(\dfrac{0,3.56}{24}.100\)=70%

=>%m Mg=100-70=30%

=>VHCl=\(\dfrac{0,3.2+0,3.2}{2}\)=0,6l=600ml

b)

 XCl₂+2AgNO₃->2AgCl+X(NO₃)₂

0,6--------------------1,2mol

=>m AgCl=1,2.143,5=172,2g

\(n_{HCl}=2.0,4=0,8(mol)\\ n_{Fe}=x(mol);n_{Al}=y(mol)\\ \Rightarrow 56x+27y=11(1)\\ Fe+2HCl\to FeCl_2+H_2\\ 2Al+6HCl\to 2AlCl_3+3H_2\\ \Rightarrow 2x+3y=0,8(2)\\ (1)(2)\Rightarrow x=0,1(mol);y=0,2(mol)\)

\(a,\Sigma n_{H_2}=x+1,5y=0,4(mol)\\ \Rightarrow V_{H_2}=0,4.22,4=8,96(l)\\ b,m_{Fe}=0,1.56=5,6(g);m_{Al}=0,2.27=5,4(g)\\ c,m_{dd_{HCl}}=400.1,12=448(g)\\ n_{FeCl_2}=0,1(mol);n_{AlCl_3}=0,2(mol)\\ \Rightarrow C\%_{FeCl_2}=\dfrac{0,1.127}{5,6+448-0,1.2}.100\%=2,8\%\\ C\%_{AlCl_3}=\dfrac{0,2.133,5}{5,4+448-0,3.2}.100\%=5,9\%\)

\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ Đặt:n_{Fe}=a\left(mol\right);n_{Zn}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}56a+65b=23,3\\a+b=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,3\\b=0,1\end{matrix}\right.\\ \Rightarrow\%m_{Zn}=\dfrac{0,1.65}{23,3}.100\approx27,897\%\\ \Rightarrow\%m_{Fe}\approx72,103\%\)

Fe+2HCl--->FeCl2+H2
    Zn+2HCl-->ZnCl2+H2
Gọi số mol của Fe và Zn lần lượt là x,y mol
=> ta có hpt {56x+65y=23,3
                     {x+y=8,96/22,4
<=>{x=0,3=>mFe=16,8g
      {y=0,1=>mZn=6,5g
nHCl=2nH2=2.8,96/22,4=0,8 mol
=>mHCl=29,2g
%mFe=16,8/23,3.100=72,10300429%
=>%mZn=27,89699571%

Chúc bn học giỏi

Gọi số mol Mg = x(mol); số mol Fe = y ( mol) 

Bảo toàn e , ta có phương trình : 

x + y = 0,03 (1)

Theo đề bài : 

24x + 56y  = 1,36 (2) 

Từ (1) và (2) ta có hệ phương trình , giải hệ ta được : 

x = 0,01 (mol) ; y = 0,02 (mol) 

=> %mMg = 0,01 . 24 / 1,36 . 100% = 17,65% 

=> %mFe = 100 - 17,65 = 82,35 %

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

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

a, Giả sử: \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)

⇒ 24x + 56y = 1,36 (1)

Ta có: \(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Mg}+n_{Fe}=x+y\left(mol\right)\)

⇒ x + y = 0,03 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,01\left(mol\right)\\y=0,02\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=\dfrac{0,01.24}{1,36}.100\%\approx17,6\%\\\%m_{Fe}\approx82,4\%\end{matrix}\right.\)

b, Theo PT: \(n_{HCl}=2n_{H_2}=0,06\left(mol\right)\)

\(\Rightarrow V_{HCl}=\dfrac{0,06}{2}=0,03\left(l\right)\)

Theo PT: \(\left\{{}\begin{matrix}n_{MgCl_2}=n_{Mg}=0,01\left(mol\right)\\n_{FeCl_2}=n_{Fe}=0,02\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{MgCl_2}}=\dfrac{0,01}{0,03}\approx0,33M\\C_{M_{FeCl_2}}=\dfrac{0,02}{0,03}\approx0,66M\end{matrix}\right.\)

Bạn tham khảo nhé!

Đặt: \(\left\{{}\begin{matrix}x=n_{Fe}\left(mol\right)\\y=n_{Al}\left(mol\right)\end{matrix}\right.\)

\(\sum m_{hh}=11\left(g\right)\Rightarrow56x+27y=11\left(1\right)\)

\(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\\ \left(mol\right)....x\rightarrow..2x........x......x\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\\ \left(mol\right)....y\rightarrow..3y.........y......1,5y\)

\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ \Rightarrow x+1,5y=0,4\left(2\right)\)

\(\xrightarrow[\left(2\right)]{\left(1\right)}\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

a) \(\%m_{Fe}=\dfrac{56.0,1}{11}=51\%\)

\(\rightarrow\%m_{Al}=49\%\)

b) \(\sum m_{ctHCl}=\left(2.0,1+3.0,2\right).36,5=29,2\left(g\right)\)

\(m_{ddHCl}=\dfrac{29,2.100\%}{10\%}=292\left(g\right)\)

c) \(m_{H_2\uparrow}=\left(1.0,1+1,5.0,2\right).2=0,8\left(g\right)\)

\(m_{ddsaupu}=m_{hh}+m_{ddHCl}-m_{H_2\uparrow}=11+292-0,8=302,2\left(g\right)\)

\(C\%_{FeCl_2}=\dfrac{0,1.127}{302,2}.100=4,2\%\\ C\%_{AlCl_3}=\dfrac{0,2.133,5}{302,2}.100=8,8\%\)

Đặt: {x=nFe(mol)y=nAl(mol){x=nFe(mol)y=nAl(mol)

∑mhh=11(g)⇒56x+27y=11(1)∑mhh=11(g)⇒56x+27y=11(1)

%mFe=56.0,111=51%%mFe=56.0,111=51%

→%mAl=49%→%mAl=49%

b) ∑mctHCl=(2.0,1+3.0,2).36,5=29,2(g)∑mctHCl=(2.0,1+3.0,2).36,5=29,2(g)