a) Do dd sau pư có 3 chát tan với nồng độ % bằng nhau

=> \(m_{Al_2\left(SO_4\right)_3}=m_{ZnSO_4}=m_{H_2SO_4\left(dư\right)}\)

Gọi số mol Al, Zn là a, b (mol)

PTHH: 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

             a----->1,5a------->0,5a----->1,5a

            Zn + H₂SO₄ --> ZnSO₄ + H₂

             b----->b--------->b----->b

=> \(\left\{{}\begin{matrix}m_{Al_2\left(SO_4\right)_3}=342.0,5a=171a\left(g\right)\\m_{ZnSO_4}=161b\left(g\right)\end{matrix}\right.\)

=> 171a = 161b 

=> \(\dfrac{a}{b}=\dfrac{161}{171}\) (1)

Có: \(\dfrac{m_{Al}}{m_{Zn}}=\dfrac{27.n_{Al}}{65.n_{Zn}}=\dfrac{27}{65}.\dfrac{161}{171}=\dfrac{483}{1235}\)

b) \(n_{H_2}=1,5a+b=\dfrac{11,2}{22,4}=0,5\left(mol\right)\) (2)

(1)(2) => \(\left\{{}\begin{matrix}a=\dfrac{161}{825}\left(mol\right)\\b=\dfrac{57}{275}\left(mol\right)\end{matrix}\right.\)

=> \(x=\dfrac{161}{825}.27+\dfrac{57}{275}.65=\dfrac{5154}{275}\left(g\right)\)

\(m_{H_2SO_4\left(dư\right)}=m_{Al_2\left(SO_4\right)_3}=342.0,5\dfrac{161}{825}=\dfrac{9177}{275}\left(g\right)\)

=> \(m_{H_2SO_4\left(bđ\right)}=98\left(1,5a+b\right)+\dfrac{9177}{275}=\dfrac{22652}{275}\left(g\right)\)

=> \(y=\dfrac{\dfrac{22652}{275}.100}{10}=\dfrac{45304}{55}\left(g\right)\)

a) Do dd sau pư có 3 chát tan với nồng độ % bằng nhau

=> \(m_{Al_2\left(SO_4\right)_3}=m_{ZnSO_4}=m_{H_2SO_4\left(dư\right)}\)

Gọi số mol Al, Zn là a, b (mol)

PTHH: 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

             a----->1,5a------->0,5a----->1,5a

            Zn + H₂SO₄ --> ZnSO₄ + H₂

             b----->b--------->b----->b

=> \(\left\{{}\begin{matrix}m_{Al_2\left(SO_4\right)_3}=342.0,5a=171a\left(g\right)\\m_{ZnSO_4}=161b\left(g\right)\end{matrix}\right.\)

=> 171a = 161b 

=> \(\dfrac{a}{b}=\dfrac{161}{171}\) (1)

Có: \(\dfrac{m_{Al}}{m_{Zn}}=\dfrac{27.n_{Al}}{65.n_{Zn}}=\dfrac{27}{65}.\dfrac{161}{171}=\dfrac{483}{1235}\)

b) \(n_{H_2}=1,5a+b=\dfrac{11,2}{22,4}=0,5\left(mol\right)\) (2)

(1)(2) => \(\left\{{}\begin{matrix}a=\dfrac{161}{825}\left(mol\right)\\b=\dfrac{57}{275}\left(mol\right)\end{matrix}\right.\)

=> \(x=\dfrac{161}{825}.27+\dfrac{57}{275}.65=\dfrac{5154}{275}\left(g\right)\)

\(m_{H_2SO_4\left(dư\right)}=m_{Al_2\left(SO_4\right)_3}=342.0,5\dfrac{161}{825}=\dfrac{9177}{275}\left(g\right)\)

=> \(m_{H_2SO_4\left(bđ\right)}=98\left(1,5a+b\right)+\dfrac{9177}{275}=\dfrac{22652}{275}\left(g\right)\)

=> \(y=\dfrac{\dfrac{22652}{275}.100}{10}=\dfrac{45304}{55}\left(g\right)\)

mn giải giiup vs ạ

Câu 5:

Ta có: \(n_{H_2}=\dfrac{14,874}{24,79}=0,6\left(mol\right)\)

Gọi: n_A = 3x (mol) ⇒ n_B = 2x (mol)

PT: \(A+2HCl\rightarrow AlCl_2+H_2\)

\(2B+6HCl\rightarrow2BCl_3+3H_2\)

Theo PT: \(n_{H_2}=n_A+\dfrac{3}{2}n_B\) 

⇒ 0,6 = 3x + 3/2.2x 

⇒ x = 0,1 (mol)

⇒ n_A = 0,3 (mol), n_B = 0,2 (mol)

Mà: m_A + m_B = 24,9

⇒ 0,3.M_A + 0,2.M_B = 24,9

\(\Rightarrow M_A=\dfrac{24,9-0,2M_B}{0,3}>60\)

⇒ M_B < 34,5 (g/mol) → M_B = 27 (g/mol) → Al

M_A = 65 (g/mol) → Zn

Câu 6:

a, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

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

Gọi: \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Zn}=y\left(mol\right)\end{matrix}\right.\) ⇒ 24x + 65y = 19,85 (1)

Theo PT: \(\left\{{}\begin{matrix}n_{MgCl_2}=n_{Mg}=x\left(mol\right)\\n_{ZnCl_2}=n_{Zn}=y\left(mol\right)\end{matrix}\right.\) 

\(n_{HCl\left(pư\right)}=2n_{Mg}+2n_{Zn}=2x+2y\left(mol\right)\)

⇒ n_{HCl (dư)} = (2x + 2y).20% (mol)

⇒ 95x + 136y + (2x + 2y).20%.36,5 = 54,09 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,15\left(mol\right)\\y=0,25\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,15.24}{19,85}.100\%\approx18,14\%\\\%m_{Zn}\approx81,86\%\end{matrix}\right.\)

b, Ta có: n_{HCl (pư)} = 0,15.2 + 0,25.2 = 0,8 (mol) ⇒ n_H2 = 1/2n_HCl = 0,4 (mol)

n_{HCl (dư)} = 0,8.20% = 0,16 (mol)

\(\Rightarrow m_{ddHCl}=\dfrac{\left(0,8+0,16\right).36,5}{29,2\%}=120\left(g\right)\)

⇒ m _{dd sau pư} = 19,85 + 120 - 0,4.2 = 139,05 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{HCl}=\dfrac{0,16.36,5}{139,05}.100\%\approx4,20\%\\C\%_{MgCl_2}=\dfrac{0,15.95}{139,05}.100\%\approx10,25\%\\C\%_{ZnCl_2}=\dfrac{0,25.136}{139,05}.100\%\approx24,45\%\end{matrix}\right.\)

Câu 7:

Ta có: 56n_Fe + M_M.n_M = 12

- Cho pư với HCl 

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(M+2HCl\rightarrow MCl_2+H_2\)

Theo PT: \(n_{H_2}=n_{Fe}+n_M=\dfrac{2,479}{24,79}=0,1\left(mol\right)\left(2\right)\)

- Cho pư với Cl₂:

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

\(M+Cl_2\underrightarrow{t^o}MCl_2\)

Theo PT: \(n_{Cl_2}=\dfrac{3}{2}n_{Fe}+n_M=\dfrac{6,1975}{24,79}=0,25\left(mol\right)\left(3\right)\)

Từ (2) và (3) ⇒ n_M = -0,2 → vô lý

→ M không pư với dd HCl.

⇒ n_Fe = n_H2 = 0,1 (mol)

Thay vào (3) ⇒ n_M = 0,1 (mol)

Thay vào (1) ⇒ M_M = 64 (g/mol)

→ M là Cu.

Gọi \(\left\{{}\begin{matrix}n_{Al}=3a\left(mol\right)\\n_R=2a\left(mol\right)\end{matrix}\right.\)

=> 81a + 2a.M_R = 12,9 (1)

PTHH: 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

             3a------------------------------>4,5a

            R + H₂SO₄ --> RSO₄ + H₂

           2a----------------------->2a

=> \(6,5a=\dfrac{1,3}{2}=0,65\)

=> a = 0,1 (mol)

=> M_R = 24 (g/mol)

=> R là Mg(Magie)