\(n_{H_2}=\frac{10,08}{22,4}=0,45\left(mol\right)\)

Ta có: \(n_{HCl\left(pư\right)}=2.n_{H_2}=0,9\left(mol\right)\)

=> \(n_{Cl}=0,9\left(mol\right)\)

Có:

\(m_{muối}=m_{hhKL}+m_{Cl}\)

= 13,1 + 0,9.35,5

= 45,05 (g)

Đáp án A

\(n_{H_2}=\dfrac{11.2}{22.4}=0.5\left(mol\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

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

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

\(n_{HCl}=2n_{H_2}=0.5\cdot2=1\left(mol\right)\)

\(BTKL:\)

\(m_X+m_{HCl}=m_M+m_{H_2}\)

\(\Rightarrow m_M=13.4+1\cdot36.5-0.5\cdot2=48.94\left(g\right)\)

\(2Al + 6HCl \to 2AlCl_3 + 3H_2\\ Fe +2HCl \to FeCl_2 + H_2\\ Mg + 2HCl \to MgCl_2 + H_2\\ n_{Cl} = n_{HCl} = 2n_{H_2} = 2.\dfrac{11,2}{22,4} = 1(mol)\\ m_{muối} = m_{kim\ loại} + m_{Cl} = 13,4 + 1.35,5 = 48,9(gam)\)

\(Đặt:n_{Mg}=a\left(mol\right);n_{Al}=b\left(mol\right)\left(a,b>0\right)\\ PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ \Rightarrow\left\{{}\begin{matrix}24a+27b=5,1\\22,4a+22,4.1,5.b=5,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\\ a,\Rightarrow\%m_{Mg}=\dfrac{0,1.24}{5,1}.100\approx47,059\%\\ \Rightarrow\%m_{Al}\approx100\%-47,059\%\approx52,941\%\\ b,n_{HCl}=2.n_{H_2}=2.\left(0,1+0,1.1,5\right)=0,5\left(mol\right)\\ \Rightarrow V_{ddHCl}=\dfrac{0,5}{2}=0,25\left(l\right)\)

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

   x           2x            x             x

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

y            3y          y             1,5y

Ta có hệ:

\(\left\{{}\begin{matrix}24x+27y=5,1\\x+1,5y=\dfrac{5,6}{22,4}=0,25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)

\(\%m_{Mg}=\dfrac{0,1\cdot24}{5,1}\cdot100\%=47,06\%\)

\(\%m_{Al}=100\%-47,06\%=52,94\%\)

b)\(\Sigma n_{HCl}=2x+3y=2\cdot0,1+3\cdot0,1=0,5mol\)

\(V=\dfrac{n}{C_M}=\dfrac{0,5}{2}=0,25l=250ml\)

Gọi số mol H₂ sinh ra là a (mol)

=> n_HCl = 2a (mol)

Theo ĐLBTKL: m_{kim loại} + m_HCl = m_muối + m_H2

=> 17,5 + 36,5.2a = 31,7 + 2a

=> a = 0,2 (mol)

=> V = 0,2.22,4 = 4,48 (l)

m_Cl^-=m_A-m_KL=14,2g⇒nCl^-=0,4⇒nH₂=0,2(mol)⇒V=0,2.22,4=4,48(l)

\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)\)

\(n_{HCl}=0.35\cdot1=0.35\left(mol\right)\)

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

Lập tỉ lệ : 

\(0.1< \dfrac{0.35}{2}\Rightarrow HCldư\)

\(m_{FeCl_2}=0.1\cdot127=12.7\left(g\right)\)

Ta có: \(\left\{{}\begin{matrix}n_{HCl}=0,796.0,5=0,398\left(mol\right)\\n_{H_2SO_4}=0,796.0,75=0,597\left(mol\right)\end{matrix}\right.\)

\(n_{H_2}=\dfrac{4,368}{22,4}=0,195\left(mol\right)\)

BTNT H, có: \(n_{HCl}+2n_{H_2SO_4}=2n_{H_2}+2n_{H_2O}\Rightarrow n_{H_2O}=0,601\left(mol\right)\)

Theo ĐLBT KL, có: m _hh + m _axit = m _muối + m_H2 + m_H2O

⇒ m = m _muối = 26,43 + 0,398.36,5 + 0,597.98 - 0,195.2 - 0,601.18 = 88,255 (g)

a) Gọi số mol Mg, Al, Fe trong m gam hỗn hợp là a, b, c (mol)

\(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)

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

_______a--------------------->a------->a_______(mol)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

_b-------------------->b------->1,5b___________(mol)

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

_c------------------>c------->c_______________(mol)

=> \(\left\{{}\begin{matrix}a+1,5b+c=0,35\left(1\right)\\95a+133,5b+127c=35,55\left(2\right)\end{matrix}\right.\)

Mặt khác:

PTHH: \(Mg+Cl_2\underrightarrow{t^o}MgCl_2\)

_______a--------------->a_________(mol)

\(2Al+3Cl_2\underrightarrow{t^o}2AlCl_3\)

_b----------------->b______________(mol)

\(2Fe+3Cl_2\underrightarrow{t^o}2FeCl_3\)

_c------------------>c______________(mol)

=> 95a + 133,5b + 162,5 = 39,1 (3)

(1)(2)(3) => \(\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,1\left(mol\right)\\c=0,1\left(mol\right)\end{matrix}\right.\) 

=> m = 24.0,1 + 27.01 + 56.0,1 = 10,7(g)

b) \(\left\{{}\begin{matrix}m_{Mg}=24.0,1=2,4\left(g\right)\\m_{Al}=27.0,1=2,7\left(g\right)\\m_{Fe}=56.0,1=5,6\left(g\right)\end{matrix}\right.\)