a) \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

Gọi số mol Zn, Al là a, b

=> 65a + 27b = 18,4 (1)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

           a----->2a------------->a

            2Al + 6HCl --> 2AlCl₃ + 3H₂

            b---->3b------------->1,5b

=> a + 1,5b = 0,5 (2)

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

=> \(\left\{{}\begin{matrix}m_{Zn}=0,2.65=13\left(g\right)\\m_{Al}=0,2.27=5,4\left(g\right)\end{matrix}\right.\)

b) n_HCl(pư) = 2a + 3b = 1 (mol)

n_HCl(dư) = 0,6.2 - 1 = 0,2 (mol)

PTHH: KOH + HCl --> KCl + H₂O

           0,2<----0,2

=> \(V=\dfrac{0,2}{1}=0,2\left(l\right)\)

\(n_{HCl}=0,6.2=1,2\left(mol\right)\\ n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\ a,n_{HCl\left(dư\right)}=1,2-2.n_{H_2}=1,2-2.0,5=0,2\left(mol\right)\\PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Đặt:n_{Zn}=a\left(mol\right);n_{Al}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}65a+27b=18,4\\a+1,5b=0,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,2\end{matrix}\right.\\ \Rightarrow m_{Zn}=0,2.65=13\left(g\right);m_{Al}=0,2.27=5,4\left(g\right)\\ b,KOH+HCl_{dư}\rightarrow KCl+H_2O\\ n_{KOH}=n_{HCl\left(dư\right)}=0,2\left(mol\right)\\ \Rightarrow V=V_{ddKOH}=\dfrac{0,2}{1}=0,2\left(l\right)\)

MnO₂+4HCl->MnCl₂+Cl₂+2H₂O

               0,5--------------0,125 mol

n _HCl =\(\dfrac{18,25}{36,5}\)=0,5 mol

=>V_Cl2=0,125.22,4=2,8l

\(n_{Cl_2}=\dfrac{20,16}{22,4}=0,9\left(mol\right)\\ Đặt:a=n_{Fe};b=n_{Al}\left(a,b>0\right)\\ a,PTHH:2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ 2Al+3Cl_2\rightarrow\left(t^o\right)2AlCl_3\\ \Rightarrow\left\{{}\begin{matrix}56a+27b=11\\1,5a+1,5b=0,9\end{matrix}\right.\)

Bấm hệ ra nghiệm âm. Nên là mình sai đề rồi em nha :D

2Fe+3Cl₂-to>2FeCl₃

x-----3\2x

2Al+3Cl₂-to>2AlCl₃

y-------3\2y

=> ta có :\(\left\{{}\begin{matrix}56x+27y=11\\\dfrac{3}{2}x+\dfrac{3}{2}y=0,9\end{matrix}\right.\)

=>Tạo ra nghiệm âm , vui lòng kiểm tra lại

\(CuO+2HCl\rightarrow CuCl_2+H_2O\\ FeO+2HCl\rightarrow FeCl_2+H_2O\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\\ Fe_3O_4+8HCl\rightarrow2FeCl_3+FeCl_2+4H_2O\)

2HCl+CuO->CuCl₂+H₂O

FeO+2HCl->FeCl₂+H₂O

Fe₂O₃+6HCl->2FeCl₃+3H₂O

Fe₃O₄+8HCl->2FeCl₃+FeCl₂+4H₂O

\(Cu\left(OH\right)_2+2HCl\rightarrow CuCl_2+2H_2O\)

\(Fe\left(OH\right)_3+3HCl\rightarrow FeCl_3+3H_2O\)

\(Fe\left(OH\right)_2+2HCl\rightarrow FeCl_2+2H_2O\)

\(2HCl+Cu\left(OH\right)_2\rightarrow CuCl_2+2H_2O\\ 3HCl+Fe\left(OH\right)_3\rightarrow FeCl_3+3H_2O\\ 2HCl+Fe\left(OH\right)_2\rightarrow FeCl_2+2H_2O\)

\(2HCl+CaCO_3\rightarrow CaCl_2+CO_2\uparrow+H_2O\\ HCl+AgNO_3\rightarrow AgCl\downarrow+HNO_3\\ HCl+NaHCO_3\rightarrow NaCl+CO_2\uparrow+H_2O\)

CaCO₃+2HCl->CaCl₂+H₂O+CO₂

AgNO₃+HCl->AgCl+HNO₃

NaHCO₃+HCl->NaCl+H₂O+CO₂

nhưng đề không nói á :v

4,8l hay 4,48l :) ?

Quy đổi X thành \(\left\{{}\begin{matrix}Mg:a\left(mol\right)\\Ca:b\left(mol\right)\\O:c\left(mol\right)\end{matrix}\right.\)

Bảo toàn Mg: \(n_{Mg}=n_{MgCl_2}=\dfrac{24,7}{95}=0,26\left(mol\right)\)

Có: 24a + 40b + 16c = 21,44

=> 40b + 16c = 21,44 - 24.0,26 = 15,2 (1)

\(n_{H_2}=\dfrac{6,496}{22,4}=0,29\left(mol\right)\)

Mg⁰ - 2e --> Mg⁺²

0,26->0,52

Ca⁰ - 2e --> Ca⁺²

b--->2b

O⁰ + 2e --> O^-2

c--->2c

2H⁺ + 2e --> H₂⁰

       0,58<--0,29

Bảo toàn e: 0,52 + 2b = 2c + 0,58

=> 2b - 2c = 0,06 (2)

(1)(2) => a = 0,28; b = 0,25

=> \(n_{CaCl_2}=n_{Ca}=0,28\left(mol\right)\)

=> \(m_{CaCl_2}=0,28.111=31,08\left(g\right)\)

- Vì hỗn hợp rắn qua dd HCl có thấy H₂ nên hh rắn chắc chắn có Al dư

\(PTHH:\left(a\right)4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ \left(b\right)Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\\ \left(c\right)2Al_{dư}+6HCl\rightarrow2AlCl_3+3H_2\\ n_{H_2\left(c\right)}=\dfrac{1,68}{22,4}=0,075\left(mol\right)\\ n_{AlCl_3\left(tổng\right)}=\dfrac{13,35}{133,5}=0,1\left(mol\right)\\ n_{AlCl_3\left(c\right)}=n_{Al\left(dư\right)}=\dfrac{2}{3}.n_{H_2\left(c\right)}=\dfrac{2.0,075}{3}=0,05\left(mol\right)\\ \Rightarrow n_{AlCl_3\left(b\right)}=0,1-0,05=0,05\left(mol\right)\\ \Rightarrow n_{HCl}=\dfrac{6}{2}.n_{AlCl_3\left(tổng\right)}=\dfrac{6}{2}.0,1=0,3\left(mol\right)\\ \Rightarrow V=V_{ddHCl}=\dfrac{0,3}{1}=0,3\left(l\right)\\ n_{Al_2O_3}=\dfrac{n_{AlCl_3\left(b\right)}}{2}=\dfrac{0,05}{2}=0,025\left(mol\right)\)

\(\Rightarrow n_{Al\left(a\right)}=2.n_{Al_2O_3}=2.0,025=0,05\left(mol\right)\\ \Rightarrow n_{Al\left(tổng\right)}=n_{Al\left(a\right)}+n_{Al\left(c\right)}=0,05+0,05=0,1\left(mol\right)\\ \Rightarrow m=m_{Al\left(tổng\right)}=0,1.27=2,7\left(g\right)\\ n_{O_2}=\dfrac{3}{2}.n_{Al_2O_3}=\dfrac{3}{2}.0,025=0,0375\left(mol\right)\\ \Rightarrow V=V_{O_2\left(đktc\right)}=0,0375.22,4=0,84\left(l\right)\)

Lần sau  bạn phải viết bài mấy vô nha, với cả bài 2 này bạn chưa làm câu c hử