\(n_{Zn} = a(mol) ; n_{Al} = b(mol) ; n_{Mg} = c(mol)\\ \Rightarrow 65a + 27b + 24c = 44,1(1)\\ Zn + 2HCl \to ZnCl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + 3 H_2\\ Mg + 2HCl \to MgCl_2 + H_2\\ n_{H_2} = a + 1,5b + c = \dfrac{31,36}{22,4} = 1,4(2)\\ Mà : 2a = 3b(3)\\ (1)(2)(3) \Rightarrow a = 0,3 ; b = 0,2 ; c = 0,8\\ \%m_{Zn} = \dfrac{0,3.65}{44,1}.100\% = 44,22\%\\ \%m_{Al} = \dfrac{0,2.27}{44,1}.100\% = 12,24\%\)

\(\%m_{Mg} = 100\% -44,22\% -12,24\% = 43,54\%\)

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)

\(n_{CaCO_3}=a\left(mol\right),n_{K_2SO_3}=b\left(mol\right)\)

\(m_{hh}=100a+158=70.3\left(g\right)\left(1\right)\)

\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

\(K_2SO_3+2HCl\rightarrow2KCl+SO_2+H_2O\)

\(n_{khí}=a+b=0.5\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.15,b=0.35\)

\(m_{Muối}=0.15\cdot111+0.35\cdot2\cdot74.5=68.8\left(g\right)\)

a) PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)  (1)

                \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)  (2)

b) Dựa vào đề, ta thấy chắc chắn HCl dư

Ta có: \(\Sigma n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Gọi số mol của Fe là \(a\) \(\Rightarrow n_{H_2\left(1\right)}=a\)

Gọi số mol của Mg là \(b\) \(\Rightarrow n_{H_2\left(2\right)}=b\)

Ta lập được hệ phương trình:

\(\left\{{}\begin{matrix}56a+24b=8\\a+b=0,2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=56\cdot0,1=5,6\left(g\right)\\m_{Mg}=24\cdot0,1=2,4\left(g\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{5,6}{8}\cdot100\%=70\%\\\%m_{Mg}=30\%\end{matrix}\right.\)

c) Theo các PTHH: \(n_{FeCl_2}=n_{MgCl_2}=n_{Fe}=n_{Mg}=0,1mol\)

\(\Rightarrow\left\{{}\begin{matrix}m_{FeCl_2}=0,1\cdot127=12,7\left(g\right)\\m_{MgCl_2}=0,1\cdot95=9,5\left(g\right)\end{matrix}\right.\) \(\Rightarrow m_{muối}=22,2\left(g\right)\)

d) Ta có: \(\Sigma n_{HCl}=\dfrac{500\cdot16\%}{36,5}=\dfrac{160}{73}\left(mol\right)\)

\(\Rightarrow n_{HCl\left(dư\right)}=\dfrac{654}{365}\left(mol\right)\) \(\Rightarrow m_{HCl\left(dư\right)}=\dfrac{654}{365}\cdot36,5=65,4\left(g\right)\)

Mặt khác: \(m_{dd}=m_{hh}+m_{ddHCl}-m_{H_2}=507,6\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{12,7}{507,6}\cdot100\%\approx2,5\%\\C\%_{MgCl_2}=\dfrac{9,5}{507,6}\cdot100\%\approx1,87\%\\C\%_{HCl\left(dư\right)}=\dfrac{65,4}{507,6}\cdot100\%\approx12,88\%\end{matrix}\right.\)

a, PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

b, Gọi: \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Mg}=y\left(mol\right)\end{matrix}\right.\) ⇒ 27x + 24y = 7,8 (1)

Ta có: m _{dd tăng} = m_KL - m_H2 ⇒ m_H2 = 7,8 - 7 = 0,8 (g) 

\(\Rightarrow n_{H_2}=\dfrac{0,8}{2}=0,4\left(mol\right)\)

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Mg}=\dfrac{3}{2}x+y=0,4\left(mol\right)\left(2\right)\)

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

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2.27}{7,8}.100\%\approx69,23\%\\\%m_{Mg}\approx30,77\%\end{matrix}\right.\)

a/ Gọi số mol Al, Mg trong hỗn hợp là a, b

PTHH:

2Al + 6HCl ===> 2AlCl₃ + 3H₂

a............................................1,5a

Mg + 2HCl ===> MgCl₂ + H₂

b.........................................b

n_H2 = 5,6 / 22,4 = 0,25 (mol)

The đề ra, ta có hệ phương trình:

\(\begin{cases}27a+24b=5,1\\1,5a+b=0,25\end{cases}\)=> \(\begin{cases}a=0,1\\b=0,1\end{cases}\)

=> m_Al = 0,1 x 27 = 2,7 gam

m_Mg = 0,1 x 24 = 2,4 gam

=> %m_Al = \(\frac{2,7}{5,1}.100\%=52,94\%\)

%m_Mg = 100% - 52,94% = 46,06%

b/ Tổng số mol của HCl = 0,3 + 0,2 = 0,5 mol

=> m_HCl = 0,5 x 36,5 = 18,25 gam

c/ Áp dụng định luật bảo toàn khối lượng, ta có
m_{hỗn hợp muối} = m_{kim loại} + m_HCl - m_H2

₌ 5,1 + 18,25 - 0,25 x 2 = 22,85 gam

n_H2= \(\dfrac{0,896}{22,4}\) = 0,04(mol)

Mg + 2HCl  \(\rightarrow\) MgCl₂  + H₂\(\uparrow\)

a                                  \(\rightarrow\)    a     (mol)

2Al + 6HCl \(\rightarrow\) 2AlCl₃ + 3H₂\(\uparrow\)         (ai dạy Al hóa trị II thế =.=)

b                                  \(\rightarrow\) 1,5b   (mol)

Gọi a,b lần lượt là số mol của Mg và Al

Theo đầu bài, ta có:   \(\left\{{}\begin{matrix}24a+27b=0.78\\a+1,5b=0,04\end{matrix}\right.\) \(\Rightarrow\)\(\left\{{}\begin{matrix}a=0,01\\b=0,02\end{matrix}\right.\)

=> m_Mg= 0,01.24 = 0,24(g)

=> m_Al = 0,78 - 0,24 = 0,54(g)