Gọi n_Al = n_Zn = a (mol)

\(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

             a---------------------->1,5a

             Zn + 2HCl --> ZnCl₂ + H₂

             a--------------------->a

=> 1,5a + a = 0,6

=> a = 0,24 (mol)

=> m_hh = 0,24.27 + 0,24.65 = 22,08 (g)

Gọi \(n_{Al}=n_{Zn}=a\left(mol\right)\)

PTHH:

2Al + 6HCl ---> 2AlCl₃ + 3H₂

a                                       a

Zn + 2HCl ---> ZnCl₂ + H₂

a                                     a

\(\rightarrow22,4\left(a+a\right)=13,44\\ \Leftrightarrow a=0,3\left(mol\right)\\ \rightarrow\left\{{}\begin{matrix}m_{Al}=0,3.27=8,1\left(g\right)\\m_{Zn}=0,3.65=19,5\left(g\right)\end{matrix}\right.\\ \rightarrow m_{hh}=8,1+19,5=27,6\left(g\right)\)

Đặt n_Al = a (mol)

Vì theo đề: n_Al = n_Zn

=> n_Zn = a (mol)

PT: 2Al + 6HCl -----> 2AlCl₃ + 3H₂\(\uparrow\)

Zn + 2HCl -----> ZnCl₂ + H₂\(\uparrow\)

Mà V_{H2 thu được} = 13,44 (l)

hay : \(\left(\dfrac{3a}{2}+a\right).22,4=13,44\Rightarrow a=0,24\)

Thay a = 0,24 vào hai PTHH, ta được

+) n_{HCl đã dùng (TT)} = \(\left(\dfrac{0,24.6}{2}+\dfrac{0,24.2}{1}\right).\dfrac{100}{80}=1,5\left(mol\right)\)

=> m_{HCl đã dùng (TT)} = 1,5 . 36,5 = 54,75(g)

+) m_{kim loại đã dùng} = \(\dfrac{0,24.27.100}{80}+\dfrac{0,24.65.100}{80}=27,6\left(g\right)\)

2Al+6HCl->2AlCl₃+3H₂

x-----------------x---------\(\dfrac{3}{2}\)x

Zn+2HCl->ZnCl₂+H₂

y---------------y--------y 

Ta có :

\(\left\{{}\begin{matrix}27x+65y=24,9\\\dfrac{3}{2}x+y=0,6\end{matrix}\right.\)

=>x=0,2 mol ,y=0,3 mol

=>m _AlCl3= 0,2.133,5=26,7g

=>m _ZnCl2 =0,3.136=40,8g

=>%m_Al=\(\dfrac{0,2.27}{24,9}.100\)=21,69%

=>%m _Zn=78,31%

a) Sửa đề: dd H₂SO₄ 9,8%

Ta có: \(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\) \(\Rightarrow m_{H_2}=0,35\cdot2=0,7\left(g\right)\)

Bảo toàn nguyên tố: \(n_{H_2SO_4}=n_{H_2}=0,35\left(mol\right)\) \(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,35\cdot98}{9,8\%}=350\left(g\right)\)

\(\Rightarrow m_{dd}=m_{KL}+m_{H_2SO_4}-m_{H_2}=361,6\left(g\right)\)

b) Tương tự câu a

Trong \(20,4g\) hỗn hợp có: \(\left\{{}\begin{matrix}n_{Zn}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\\n_{Al}=c\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow65a+56b+27c=20,4\left(1\right)\)

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

\(BTe:2n_{Zn}+2n_{Fe}+3n_{Al}=2n_{H_2}\)

\(\Rightarrow2a+2b+3c=2\cdot0,45\left(2\right)\)

Trong \(0,2mol\) hhX có \(\left\{{}\begin{matrix}Zn:ka\left(mol\right)\\Fe:kb\left(mol\right)\\Al:kc\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow ka+kb+kc=0,2\)

\(n_{Cl_2}=\dfrac{6,16}{22,4}=0,275mol\)

\(BTe:2n_{Zn}+3n_{Fe}+3n_{Al}=2n_{Cl_2}\)

\(\Rightarrow2ka+3kb+3kc=2\cdot0,275\)

Xét thương:

 \(\dfrac{ka+kb+kc}{2ka+3kb+3kc}=\dfrac{0,2}{2\cdot0,275}\Rightarrow\dfrac{a+b+c}{2a+3b+3c}=\dfrac{4}{11}\)

\(\Rightarrow3a-b-c=0\left(3\right)\)

Từ (1), (2), (3)\(\Rightarrow\left\{{}\begin{matrix}a=0,1mol\\b=0,2mol\\c=0,1mol\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_{Zn}=6,5g\\m_{Fe}=11,2g\\m_{Al}=2,7g\end{matrix}\right.\)