a,\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:     0,2      0,4                       0,2

\(\Rightarrow\%m_{Zn}=\dfrac{0,2.65.100\%}{21,1}=61,61\%;\%m_{ZnO}=100-61,61=38,39\%\)

b,\(n_{ZnO}=\dfrac{21,1-13}{81}=0,1\left(mol\right)\)

PTHH: ZnO + 2HCl → ZnCl₂ + H₂O

Mol:     0,1       0,2         

\(m_{ddHCl}=\dfrac{\left(0,2+0,4\right).36,5.100\%}{7,3\%}=300\left(g\right)\)

c,

PTHH: Zn + H₂SO₄ → ZnSO₄ + H₂

Mol:     0,2       0,2

PTHH: ZnO + H₂SO₄ → ZnSO₄ + H₂O

Mol:      0,1        0,1

\(n_{H_2SO_4}=0,2+0,1=0,3\left(mol\right)\Rightarrow V_{ddH_2SO_4}=\dfrac{0,3}{0,5}=0,6\left(l\right)=600\left(ml\right)\)

\(m_{ddH_2SO_4}=600.1,12=672\left(g\right)\)

*Sửa đề: "13,44 lít H₂" và "24,9 gam hh 2 kim loại"

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

               a_____3a_____________\(\dfrac{3}{2}\)a                (mol)

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

               b_____2b_____________b                   (mol)

Ta lập HPT: \(\left\{{}\begin{matrix}27a+65b=24,9\\\dfrac{3}{2}a+b=\dfrac{13,44}{22,4}=0,6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Al}=0,2\left(mol\right)\\n_{Zn}=0,3\left(mol\right)\\n_{HCl}=1,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,2\cdot27=5,4\left(g\right)\\m_{Zn}=19,5\left(g\right)\\m_{ddHCl}=\dfrac{1,2\cdot36,5}{7,3\%}=600\left(g\right)\end{matrix}\right.\)

\(a,\) Đặt \(\begin{cases} n_{Fe}=x(mol)\\ n_{Al}=y(mol) \end{cases}\Rightarrow 56x+27y=22(1)\)

\(n_{H_2}=\dfrac{17,92}{22,4}=0,8(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ 2Al+6HCl\to 2AlCl_3+3H_2\\ \Rightarrow x+1,5y=0,8(2)\\ (1)(2)\Rightarrow \begin{cases} x=0,2(mol)\\ y=0,4(mol) \end{cases} \Rightarrow \begin{cases} \%_{Fe}=\dfrac{0,2.56}{22}.100\%=50,91\%\\ \%_{Al}=100\%-50,91\%=49,09\% \end{cases} \)

\(b,\Sigma n_{HCl}=2n_{Fe}+3n_{Al}=0,4+1,2=1,6(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{1,6.36,5}{3,7\%}=1578,38\%\)

\(a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ \Rightarrow n_{Zn}=n_{H_2}=\dfrac{3,7185}{24,79}=0.,15(mol)\\ \Rightarrow m_{Zn}=0,15.65=9,75(g)\\ \Rightarrow \%_{Zn}=\dfrac{9,75}{10}.100\%=97,5\%\\ \Rightarrow \%_{Cu}=100\%-97,5\%=2,5\%\\ b,n_{HCl}=2n_{H_2}=0,3(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{0,3.36,5}{14\%}=78,21(g)\)

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

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

b, Ta có: m _{dd tăng} = m_KL - m_H2

⇒ m_H2 = 16,6 - 15,6 = 1 (g) \(\Rightarrow n_{H_2}=\dfrac{1}{2}=0,5\left(mol\right)\)

Có: 27n_Al + 56n_Fe = 16,6 (1)

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=0,5\left(2\right)\)

Từ (1) và (2) ⇒ n_Al = n_Fe = 0,2 (mol)

\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,2.27=5,4\left(g\right)\\m_{Fe}=0,2.56=11,2\left(g\right)\end{matrix}\right.\)

c, Theo PT: \(n_{HCl}=2n_{H_2}=1\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{1.36,5}{40\%}=91,25\left(g\right)\)

⇒ m _{dd sau pư} = 91,25 + 15,6 = 106,85 (g)

Theo PT: \(\left\{{}\begin{matrix}n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\\n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{AlCl_3}=\dfrac{0,2.133,5}{106,85}.100\%\approx24,98\%\\C\%_{FeCl_2}=\dfrac{0,2.127}{106,85}.100\%\approx23,77\%\end{matrix}\right.\)

\(2Al+6HCl->2AlCl_3+3H_2\\ Fe+2HCl->FeCl_2+H_2\\ n_{H_2}=\dfrac{16,6-15,6}{2}=0,5mol\\ n_{HCl}=1mol\\ n_{Al}=a;n_{Fe}=b\\ 27a+56b=16,6\\ 3a+2b=1\\ a=b=0,2\\ m_{Al}=0,2.27=5,4g\\ m_{Fe}=0,2.56=11,2g\\ m_{ddsau}=15,6+\dfrac{36,5}{0,4}=106,85g\\ C\%_{AlCl_3}=\dfrac{133,5.0,2}{106,85}.100\%=24,99\%\\ C\%_{FeCl_2}=\dfrac{127.0,2}{106,85}.100\%=23,77\%\)

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

a) Theo Pt : \(n_{H2}=n_{Fe}=n_{FeCl2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(\%m_{Fe}=\dfrac{0,1.56}{10}.100\%=56\%\)

\(\%m_{Cu}=100\%-56\%=44\%\)

b) Theo Pt : \(n_{H2}=2n_{HCl}=2.0,1=0,2\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,2.36,5}{7,3\%}.100\%=100\left(g\right)\)

c) \(m_{ddspu}=10+100-0,1.2=109,8\left(g\right)\)

\(C\%_{FeCl2}=\dfrac{0,1.127}{109,8}.100\%=11,57\%\)