Bài 1:

a, Ta có :

\(m_{Ct_{HCl}}=\frac{20}{100}.91,25=18,25\left(g\right)\)

\(\rightarrow n_{HCl}=\frac{18,25}{36,5}=0,5\left(mol\right)\)

Gọi số mol của Fe là x , số mol của Fe2O3 là y

Theo đề bài :

\(56x+160y=13,6\) (1)

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

x_____2x_______________x

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

y_______6y__________________

\(\rightarrow2x+6y=0,5\) (2)

Giải hệ phương trình (1) và (2) ta có :

\(\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)

\(\rightarrow\%_{Fe}=\frac{0,1.56}{13,6}.100\%=41,18\%\)

\(\%_{Fe2O3}=100\%-41,18\%=58,82\%\)

b, \(m_{dd_{spu}}=13,6+91,25-0,1.2=104,65\left(g\right)\)

Mà \(n_{H2}=n_{Fe}=0,1\left(mol\right)\)

Theo phương trình : \(nFeCl_2=n_{Fe}=0,1\left(mol\right)\)

\(\rightarrow\%_{FeCl2}=\frac{0,1.127}{104,65}.100\%=12,14\%\)

Theo phương trình : \(n_{FeCl3}=2n_{Fe2O3}=0,05.2=0,2\left(mol\right)\)

\(\rightarrow\%_{FeCl3}=\frac{0,1.162,5}{104,65}.100\%=15,53\%\)

Bài 2:

\(KCl+AgNO_3\rightarrow AgCl+KNO_3\)

x ________________x__________

\(NaCl+AgNO_3\rightarrow AgCl+NaNO_3\)

y _______________ y_____________

Giải hệ phương trình :

\(\left\{{}\begin{matrix}74,5x+58,5y=26,5\\143,5x+143,5y=57,4\end{matrix}\right.\rightarrow x=y=0,2\left(mol\right)\)

\(C\%_{KCl}=\frac{0,2.74,5}{500}.100\%=2,98\%\)

\(C\%_{NaCl}=\frac{0,2.58,5}{500}.100\%=2,34\%\)

Bài 3:

\(m_{dd_{HCl}}=1,19.100,8=119,952\)

\(m_{HCl}=119,952.36,5\%=43,78248\)

\(n_{HCl}=1,19976\)

\(\rightarrow n_{H2}=0,4\left(mol\right)\)

\(PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\)

_______0,4____0,8______________0,4_(mol)

\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)

0,19976____0,19952__________________

\(\rightarrow m_A=0,4.65+0,19976.\left(65+16\right)\)

\(=42,18\)

Gọi nFe = a (mol); nMg = b (mol)

56a + 24b = 8 (1)

nH2 = 4,48/22,4 = 0,2 (mol)

PTHH: 

Fe + 2HCl -> FeCl2 + H2

a ---> a ---> a ---> a

Mg + 2HCl -> MgCl2 + H2

b ---> b ---> b ---> b

a + b = 0,2 (2)

(1)(2) => a = b = 0,1 (mol)

mFe = 0,1 . 56 = 5,6 (g)

%mFe = 5,6/8 = 70%

%mMg = 100% - 70% = 30%

nHCl = 0,1 . 2 + 0,1 . 2 = 0,4 (mol)

CMddHCl = 0,4/0,1 = 4M

a.\(n_{H_2}=\dfrac{V_{H_2}}{22,4}=\dfrac{8,96}{22,4}=0,4mol\)

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

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

 x                                      x ( mol )

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

 y                                        y   ( mol )

Ta có:

\(\left\{{}\begin{matrix}56x+65y=25,55\\x+y=0,4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,05\\y=0,35\end{matrix}\right.\)

\(\Rightarrow m_{Fe}=0,05.56=2,8g\)

\(\Rightarrow m_{Zn}=0,35.65=22,75g\)

\(\%m_{Fe}=\dfrac{2,8}{25,55}.100=10,95\%\)

\(\%m_{Zn}=100\%-10,95\%=89,05\%\)

b.\(n_{HCl}=2.0,05+2.0,35=0,8mol\)

\(C_M=\dfrac{n}{V}\Rightarrow V=\dfrac{n}{C_M}=\dfrac{0,8}{2}=0,4l\)

\(a.BTNT\left(H\right):n_{HCl}=2n_{H_2}=0,65\left(mol\right)\\ \Rightarrow CM_{HCl}=\dfrac{0,65}{0,5}=1,3M\\ b.2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ Đặt:\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+y=0,325\\27x+56y=9,65\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,15\\y=0,1\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}m_{Al}=4,05\left(g\right)\\m_{Fe}=5,6\left(g\right)\end{matrix}\right.\)

\(a.Fe+2HCl\rightarrow FeCl_2+H_2\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ b.Chấtrắnkhôngtan:Cu\\ \%m_{Cu}=54,24\%\\ Đặt:\left\{{}\begin{matrix}n_{Fe}=x\left(mol\right)\\n_{Mg}=y\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}2x+2y=\dfrac{91,25.20\%}{36,5}\\56x+24y=23,6-12,8\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,15\\y=0,1\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=35,59\%\\\%m_{Mg}=10,17\%\end{matrix}\right.\)

\(a.Mg+2HCl->MgCl_2+H_2\\ Fe+2HCl->FeCl_2+H_2\\ b.Giả.sử:có:100g.dd.HCl\\ n_{HCl}=\dfrac{20\%.100}{36,5}=\dfrac{40}{73}mol\\ n_{Fe}=a;n_{Mg}=b\\ 2a+2b=\dfrac{40}{73}\\ BTKL:m_{ddsau}=56a+24b+100-2\left(a+b\right)=54a+22b+100\left(g\right)\\ C\%_{MgCl_2}=\dfrac{95b}{54a+22b+100}=\dfrac{11,787}{100}\\ -54a+783,97b=100\\ a=b=0,137\left(mol\right)\\ C\%FeCl_2=\dfrac{0,137\cdot127}{\dfrac{95\cdot0,137}{11,787\%}}\cdot100\%=15,757\%\)

a,Giả sử m_ddHCl = 36,5 (g) \(\Rightarrow n_{HCl}=\dfrac{36,5.0,2}{36,5}=0,2\left(mol\right)\)

PTHH: Fe + 2HCl → FeCl₂ + H₂

Mol:     a         2a          a         a

PTHH: Mg + 2HCl → MgCl₂ + H₂

Mol:       b         2b           b         b

Ta có: \(2a+2b=0,2\Leftrightarrow a+b=0,1\left(mol\right)\)

m_{dd D} = 56a+24b+36,5-(a+b).2 = 56a+24b+36,3 (g)

\(C\%_{ddFeCl_2}=\dfrac{127a.100\%}{56a+24b+36,3}=15,757\%\)

\(\Leftrightarrow127a=8,82392a+3,78168b+5,719791\)

\(\Leftrightarrow118,17608a-3,78168b=5,719791\)

Ta có: \(\left\{{}\begin{matrix}a+b=0,1\\118,17608a-3,78168b=5,719791\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,05\\b=0,05\end{matrix}\right.\)

\(C\%_{ddMgCl_2}=\dfrac{95.0,05.100\%}{56.0,05+24.0,05+36,3}=11,79\%\)