Có n_Zn= 0,65/65=0,1(mol)

a,PTPƯ: Zn + 2HCl ---> ZnCl₂ + H₂

(mol) 0,1 ---> 0,05 ---> 0,1 ---> 0,1

b, Theo pt có m_muối= 0,1.136=13.6(g)

Lại có V_khí=0,1.22,4= 2,24(l)

c, Theo pt có m_HCl= 0,05.36,5=1,825(g)

=> C%=\(\dfrac{1,825}{200}\).100%=0,9125%

\(n_{AlCl_3}=\dfrac{26.7}{133.5}=0.2\left(mol\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(.........0.6......0.2.......0.3\)

\(V_{H_2}=0.3\cdot22.4=6.72\left(l\right)\)

\(C\%HCl=\dfrac{0.6\cdot36.5}{300}\cdot100\%=7.3\%\)

PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

Ta có: \(n_{HCl}=\dfrac{200\cdot71\%}{36,5}=\dfrac{284}{73}\left(mol\right)\)

\(\Rightarrow n_{Fe}=n_{FeCl_2}=n_{H_2}=\dfrac{142}{73}\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=\dfrac{142}{73}\cdot56\approx108,93\left(g\right)\\m_{FeCl_2}=\dfrac{142}{73}\cdot127\approx247,04\left(g\right)\\m_{H_2}=\dfrac{142}{73}\cdot2\approx3,89\left(g\right)\\V_{H_2}=\dfrac{142}{73}\cdot22,4\approx43,57\left(l\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{Fe}+m_{ddHCl}-m_{H_2}=305,04\left(g\right)\)

\(\Rightarrow C\%_{FeCl_2}=\dfrac{247,04}{305,04}\cdot100\%\approx80,99\%\)

Fe + 2HCl ➝ FeCl₂ + H₂

mHCl = 200.71% = 142 (g) => nHCl = \(\dfrac{284}{73}\) (mol)

nFe = \(\dfrac{1}{2}\) nHCl = \(\dfrac{142}{73}\) (mol) => m ≃ 108,9 (g)

nH₂ = nFe => V ≃ 43,57 (l)

nFeCl₂ = nFe => C% ≃ 80%

(Mk nghĩ bạn nên kiểm tra lại đề vì số liệu không được đẹp cho lắm)

Ta có: \(n_{HCl}=\dfrac{200}{1000}.2=0,4\left(mol\right)\)

\(PTHH:Mg+2HCl--->MgCl_2+H_2\uparrow\left(1\right)\)

a. Theo PT₍₁₎: \(n_{Mg}=n_{H_2}=n_{MgCl_2}=\dfrac{1}{2}.n_{HCl}=\dfrac{1}{2}.0,4=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,2.24=4,8\left(g\right)\\V_{H_2}=0,2.22,4=4,48\left(lít\right)\end{matrix}\right.\)

b. \(PTHH:2NaOH+MgCl_2--->Mg\left(OH\right)_2\downarrow+2NaCl\left(2\right)\)

Ta có: \(n_{NaOH}=\dfrac{\dfrac{20\%.100}{100\%}}{40}=0,5\left(mol\right)\)

Ta thấy: \(\dfrac{0,5}{2}>\dfrac{0,2}{1}\)

Vậy NaOH dư.

Theo PT₍₂₎: \(n_{Mg\left(OH\right)_2}=n_{MgCl_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Mg\left(OH\right)_2}=0,2.58=11,6\left(g\right)\)

a: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

200ml=0,2 lít

\(n_{HCl}=0.2\cdot22.4=4.48\left(mol\right)\)

\(\Leftrightarrow n_{H_2}=2.24\left(mol\right)\)

\(\Leftrightarrow m_{H_2}=n_{H_2}\cdot M=2.24\cdot1=2.24\left(g\right)\)

\(n_{MgCl_2}=2.24\left(mol\right)\)

\(\Leftrightarrow n_{Mg}=2.24\left(mol\right)\)

\(\Leftrightarrow m_{Mg}=2.24\cdot24=53.76\left(g\right)\)

nH2 = 4.48/22.4 = 0.2 (mol) 

Zn + 2HCl => ZnCl2 + H2 

0.2......0.4.....................0.2

mZn = 65*0.2 = 13 (g) 

mHCl = 0.4*36.5 = 14.6 (g) 

C%HCl = 14.6*100%/200 = 7.3%

a)\(n_K=\dfrac{0,39}{39}=0,01mol\)

\(\left\{{}\begin{matrix}X:KOH\\Y:H_2\end{matrix}\right.\)

b)\(2K+2H_2O\rightarrow2KOH+H_2\)

   0,01      0,01         0,01     0,005

\(V_{H_2}=0,005\cdot22,4=0,112l=112ml\)

$a\big)$

$M_A=9,4.2=18,8(g/mol)$

$\to \dfrac{n_{CO_2}}{n_{H_2}}=\dfrac{18,8-2}{44-18,8}=\dfrac{2}{3}$

Mà $n_{CO_2}+n_{H_2}=\dfrac{11,2}{22,4}=0,5(mol)$

\(\begin{array} {l} \to n_{CO_2}=0,2(mol);n_{H_2}=0,3(mol)\\ Fe+2HCl\to FeCl_2+H_2\\ FeCO_3+2HCl\to FeCl_2+CO_2+H_2O\\ \text{Theo PT: }n_{Fe}=n_{H_2}=0,3(mol);n_{FeCO_3}=n_{CO_2}=0,2(mol)\\ \to m=0,3.56+0,2.116=40(g) \end{array}\)

$b\big)$

Đổi $400ml=0,4l$

\(\begin{array} {l} \text{Theo PT: }n_{FeCl_2}=n_{H_2}+n_{CO_2}=0,5(mol)\\ \to C_{M\,FeCl_2}=\dfrac{0,5}{0,4}=1,25M \end{array}\)

$c\big)$

\(\begin{array}{l} m_{dd\,FeCl_2}=\dfrac{400}{1,2}\approx 333,33(g)\\ \to C\%_{FeCl_2}=\dfrac{0,5.127}{333,33}.100\%=19,05\%\end{array}\)

 Mời các chiến thần trả lời giúp mình

`a)PTHH:`

`Mg + 2HCl -> MgCl_2 + H_2`

`0,3`      `0,6`             `0,3`       `0,3`    `(mol)`

`b) n_[Mg] = [ 7,2 ] / 24 = 0,3 (mol)`

 `=> V_[H_2] = 0,2 . 22,4 =6,72 (l)`

`c) C_[M_[HCl]] = [ 0,6 ] / [ 0,3 ] = 2 (M)`