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a) Ban đầu không hiện tượng sau một thời gian thu được kết tủa keo trắng rồi tan
$NaOH + HCl \to NaCl + H_2O$
$NaAlO_2 + HCl + H_2O \to NaCl + Al(OH)_3$
$Al(OH)_3 + 3HCl \to AlCl_3 + 3H_2O$
b)
$n_{NaOH} = 1,1(mol)$
Ta có :
$n_{HCl} = 4n_{NaAlO_2} - n_{Al(OH)_3} + n_{NaOH}$
$\Rightarrow 3,8 = 4n_{NaAlO_2} - 1,1 + 1,1$
$\Rightarrow n_{NaAlO_2} = 0,95(mol)$
Vậy :
$C_{M_{NaOH}} = \dfrac{1,1}{0,5} = 2,2M$
$C_{M_{NaAlO_2}} = \dfrac{0,95}{0,5}= 1,9M$
a) Ban đầu:
HCl + NaOH ---------> NaCl + H2O
NaAlO2 + HCl + H2O → Al(OH)3↓ + NaCl.
Sau đó, Al(OH)3 ↓ + 3HCl → AlCl3 + 3H2O
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\begin{cases} n_{Fe}=x(mol)\\ n_{Fe_2O_3}=y(mol)\\ n_{FeCO_3}=z(mol \end{cases} \Rightarrow 56x+160y+116c=30,4(1)\)
\(PTHH:Fe+2HCl\to FeCl_2+H_2\\ Fe_2O_3+6HCl\to 2FeCl_3+3H_2O\\ FeCO_3+2HCl\to FeCl_2+H_2O+CO_2\uparrow\\ M_{\text{hh khí}}=15.2=30(g/mol)\\ \Rightarrow \dfrac{m_{H_2}+m_{CO_2}}{n_{H_2}+n_{CO_2}}=30\\ \Rightarrow \dfrac{2x+44z}{x+z}=30\\ \Rightarrow 28x-14z=0(2)\)
\(m_{muối}=m_{FeCl_2}+m_{FeCl_3}=51,55\\ \Rightarrow 127(x+z)+162,5.2y=51,55\\ \Rightarrow 127x+325y+127z=51,55(3)\\ (1)(2)(3)\Rightarrow \begin{cases} x=0,05(mol)\\ y=z=0,1(mol) \end{cases} \\ \Rightarrow \begin{cases} \%_{Fe}=\dfrac{0,05.56}{30,4}.100\%=9,21\%\\ \%_{Fe_2O_3}=\dfrac{0,1.160}{30,4}.100\%=52,63\%\\ \%_{FeCO_3}=100\%-9,21\%-52,63\%=38,16\%\\ \end{cases}\\ b,\Sigma n_{HCl}=2x+6y+2z=0,9(mol)\\ \Rightarrow C_{M_{HCl}}=\dfrac{0,9}{1,8}=0,5M\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
Gọi A là kim loại hóa trị II cần tìm
\(n_{H_2}=\dfrac{0,28}{22,4}=0,0125\left(mol\right)\\ A+2HCl\rightarrow ACl_2+H_2\\ n_A=n_{H_2}=0,0125\left(mol\right)\\ M_A=\dfrac{0,3}{0,0125}=24\left(\dfrac{g}{mol}\right)\)
Vậy: Kim loại A(II) cần tìm là Magie (Mg=24)
Bài 2:
- Chỉ có Zn tác dụng với dd HCl dư chứ Cu thì không.
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{Zn}=n_{H_2}=0,2\left(mol\right)\\ \rightarrow m_{Zn}=0,2.65=13\left(g\right)\\ \rightarrow m_{Cu}=15-13=2\left(g\right)\\ Vậy:m=2\left(g\right)\)
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\(Al_2\left(SO_4\right)_3+6NaOH\rightarrow2Al\left(OH\right)_3+3Na_2SO_4\\ Al\left(OH\right)_3+NaOH\rightarrow NaAlO_2+2H_2O\\ Fe\left(OH\right)_3+3HCl\rightarrow FeCl_3+3H_2O\\ 2Fe\left(OH\right)_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+6H_2O\\ HCl+NaOH\rightarrow NaCl+H_2O\\ 2HCl+Fe\rightarrow FeCl_2+H_2\\ CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\\ CuSO_4+Fe\rightarrow FeSO_4+Cu\\ Ba\left(NO_3\right)_2+H_2SO_4\rightarrow BaSO_4+2HNO_3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a.n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ b.0,1........0,2..........0,1.........0,1\left(mol\right)\\ C\%_{ddH_2SO_4}=\dfrac{0,1.98}{140}.100=7\%\\ c.m_{dd.muối}=6,5+140-0,1.2=146,3\left(g\right)\\ C\%_{ddZnCl_2}=\dfrac{136.0,1}{146,3}.100\approx9,296\%\\ \)
a.nZn=6,565=0,1(mol)Zn+2HCl→ZnCl2+H2b.0,1........0,2..........0,1.........0,1(mol)C%ddH2SO4=0,1.98140.100=7%c.mdd.muối=6,5+140−0,1.2=146,3(g)C%ddZnCl2=136.0,1146,3.100≈9,296%
A) PTHH :
Mg + 2HCl → MgCl2 + H2
MgO + 2HCl → MgCl2 + H2O
B)
\(n_{Mg}=n_{H_2}=\dfrac{1,12}{24,79}=0,045mol\)
=> \(m_{Mg}=0,045\times24=1,1g\)
\(m_{MgO}=\)mhỗn hợp \(-m_{Mg}=9,2-1,1=8,1g\)