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

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

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

a)

$Fe + H_2SO_4 \to FeSO_4 + H_2$

b) $n_{Fe} = \dfrac{14}{56} = 0,25(mol)$
Theo PTHH : $n_{H_2SO_4} = n_{Fe} =  0,25(mol)$

$\Rightarrow m_{dd\ H_2SO_4} = \dfrac{0,25.98}{10\%} = 245(gam)$

c)

$n_{H_2} = n_{Fe} = 0,25(mol)$

Sau phản ứng, $m_{dd} = 14 + 245 - 0,25.2 = 258,5(gam)$

$C\%_{FeSO_4} = \dfrac{0,25.152}{258,5}.100\% = 14,7\%$

Khí màu nâu là NO₂

\(Fe\rightarrow Fe^{3+}+3e\)                                      \(\) \(2H^++NO_3^-+1e\rightarrow NO_2+H_2O\)

Bảo toàn e: \(n_{Fe}.3=n_{NO_2}.1\\ \Rightarrow n_{NO_2}=0,3\left(mol\right)\)

Ta có : \(n_{H^+}=n_{HNO_3}=2n_{NO_2}=0,6\left(mol\right)\\ \Rightarrow V_{HNO_3}=\dfrac{0,6}{0,5}=1,2\left(l\right)\)

Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

\(PTHH:Fe+2HNO_3--->Fe\left(NO_3\right)_2+H_2\uparrow\)

Theo PT: \(n_{HNO_3}=2.n_{Fe}=2.0,1=0,2\left(mol\right)\)

\(\Rightarrow V_{dd_{HNO_3}}=\dfrac{0,2}{0,5}=0,4\left(lít\right)\)

a) Đặt nAl=a(mol) ; nFe=b(mol) (a,b>0)

nHCl= (365.12%)/36,5=1,2(mol)

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

a________3a_________2a____1,5a(mol)

Fe + 2 HCl -> FeCl2 + H2

b_____2b____b____b(mol)

Ta có hpt:

\(\left\{{}\begin{matrix}27a+56b=22,2\\3a+2b=1,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,3\end{matrix}\right.\)

b) => %mAl= [(0,2.27)/22,2].100=24,324%

=>%mFe= 75,676%

c) mFeCl2=127. 0,3=38,1(g)

mAlCl3= 133,5. 0,2= 26,7(g)

mddsau= 22,2+365 - 1,2.2=384,8(g)

=>C%ddFeCl2= (38,1/384,8).100=9,901%

C%ddAlCl3= (26,7/384,8).100=6,939%

Câu này anh nhớ sáng em hỏi nhưng thiếu chữ HCl đúng không? Có bạn Bảo Trí làm rồi nè.

n Fe = 5,6/56 = 0,1 mol

Fe + 2HCl → FeCl2 +  H2

Theo PTHH :

n HCl = 2n Fe = 0,1.2 = 0,2(mol)

=> V dd = V dd HCl = 0,2/2 = 0,1(lít)

n FeCl2 = n Fe = 0,1(mol)

=> CM FeCl2 = 0,1/0,1 = 1M

Bài 1 : 

\(a) CaCO_3 \xrightarrow{t^o} CaO + CO_2\\ 2Fe(OH)_3 \xrightarrow{t^o} Fe_2O_3 + 3H_2O\\ b) CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O\\ 2Fe(OH)_3 + 6HCl \to 2FeCl_3 + 6H_2O\\ NaOH + HCl \to NaCl + H_2O\\ c) 2AgNO_3 + 2NaOH \to 2NaNO_3 + Ag_2O + H_2O\\ NaCl + AgNO_3 \to AgCl + NaNO_3\)

Bài 2 : 

\(a)2AgNO_3 + BaCl_2 \to 2AgCl + Ba(NO_3)_2\\ n_{AgCl} = n_{AgNO_3} = 0,2.1 = 0,2(mol)\\ \Rightarrow m_{AgCl} = 0,2.143,5 = 28,7(gam)\\ b) n_{BaCl_2} = \dfrac{1}{2}n_{AgNO_3} = 0,1(mol)\\ V_{dd\ BaCl_2} = \dfrac{0,1}{2} = 0,05(lít)\)

\(1.\)

\(a.\)

\(CaCO_3\underrightarrow{t^0}CaO+CO_2\)

\(2Fe\left(OH\right)_3\underrightarrow{t^0}Fe_2O_3+3H_2O\)

\(b.\)

\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

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

\(NaOH+HCl\rightarrow NaCl+H_2O\)

\(c.\)

\(2NaOH+2AgNO_3\rightarrow2NaNO_3+Ag_2O+H_2O\)

\(NaCl+AgNO_3\rightarrow AgCl+NaNO_3\)

\(2.\)

\(n_{AgNO_3}=0.2\cdot1=0.2\left(mol\right)\)

\(BaCl_2+2AgNO_3\rightarrow AgCl+Ba\left(NO_3\right)_2\)

\(0.1............0.2............0.1\)

\(m_{AgCl}=0.1\cdot143.5=14.35\left(g\right)\)

\(V_{dd_{BaCl_2}}=\dfrac{0.1}{2}=0.05\left(l\right)\)