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

\(n_{HCl}=0,5.1=0,5\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,5}{2}\), ta được HCl dư.

Theo PT: \(n_{HCl\left(pư\right)}=2n_{Fe}=0,2\left(mol\right)\Rightarrow n_{HCl\left(dư\right)}=0,5-0,2=0,3\left(mol\right)\)

\(\Rightarrow m_{HCl\left(dư\right)}=0,3.36,5=10,95\left(g\right)\)

b, \(n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\Rightarrow m_{FeCl_2}=0,1.127=12,7\left(g\right)\)

c, \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

d, \(m_{HCl}=0,5.36,5=18,25\left(g\right)\Rightarrow C\%_{HCl}=\dfrac{18,25}{200}.100\%=9,125\%\)

\(a.n_{Fe}=\dfrac{5,6}{56}=0,1mol\\ n_{HCl}=0,5.1=0,5mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ \Rightarrow\dfrac{0,1}{1}< \dfrac{0,5}{2}\Rightarrow HCl.dư\\ n_{HCl}=2n_{Fe}=0,2mol\\ m_{HCl\left(dư\right)}=\left(0,5-0,2\right).36,5=10,95\%\\ b)n_{Fe}=n_{FeCl_2}=n_{H_2}=0,1mol\\ m_{FeCl_2}=0,1.12,7g\\ c)V_{H_2}=0,1.22,4=2,24l\\ d)C_{\%HCl}=\dfrac{0,2.36,5}{200}\cdot100=3,65\%\)

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

\(n_{HCl}=0,5.1=0,5\left(mol\right)\)

PTHH :

\(Fr+2HCl\rightarrow FeCl_2+H_2\uparrow\)

0,1     0,2           0,1        0,1 

\(\dfrac{0,1}{1}< \dfrac{0,5}{2}\)

--> Tính theo Fe 

HCl dư 

\(a,m_{HCldư}=\left[0,5.\left(0,1.2\right)\right].36,5=10,95\left(g\right)\)

\(b,m_{FeCl_2}=0,1.127=12,7\left(g\right)\)

\(c,V_{H_2}=0,1.22,4=2,24\left(l\right)\)

\(d,m_{HCl}=0,5.36,5=18,25\left(g\right)\)

\(C\%_{HCl}=\dfrac{18,25}{200}.100\%=9,125\%\)

\(n_{Zn}=\dfrac{m}{M}=\dfrac{13}{65}=0,2\left(mol\right)\)

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

                   1          2               1            1

                  0,2       0,4            0,2         0,2 (mol)

\(a)V_{H_2}=n\cdot22,4=0,2\cdot22,4=4,48\left(l\right)\)

\(b)m_{ZnCl_2}=n\cdot M=0,2\cdot\left(65+35,5\cdot2\right)=27,2\left(g\right)\)

\(c)400ml=0,4l\\ C_{M_{HCl}}=\dfrac{n}{V_{dd}}=\dfrac{0,4}{0,4}=1M.\)

a) PTHH:

Zn + 2HCl -> ZnCl2 + H2

0,2     0,4         0,2       0,2

nZn = 13 : 65 = 0,2 (mol)

VH2 = 0,2 . 22,4 = 4,48 (l)

b) mZnCl2 = 0,2 . 201 = 40,2 (g)

c) CM HCl = 0,4 : 0,4 = 1 (M)

Số mol SO₂ và NaOH lần lượt là 0,2 và 0,25.

1 < OH^-/SO₂=1,25 < 2 ⇒ Dung dịch X chứa hai muối Na₂SO₃ và NaHSO₃.

\(n_{Na_2SO_3}=0,25-0,2=0,05\left(mol\right)\) ⇒ \(n_{NaHSO_3}=0,2-0,05=0,15\left(mol\right)\).

1. Khối lượng muối có trong X:

m=0,05.126+0,15.104=21,9 (g).

2. Nồng độ mol/l các chất trong X:

\(C_{M\left(Na_2SO_3\right)}\)=0,05/0,2=0,25 (mol/l).

\(C_{M\left(NaHSO_3\right)}\)=0,15/0,2=0,75 (mol/l).

3. Khối lượng kết tủa BaSO₃ là:

m'=0,2.217=43,4 (g).

​​

a) PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

                \(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)

b) Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)=n_{Zn}\) \(\Rightarrow n_{ZnO}=\dfrac{20-0,1\cdot65}{81}=\dfrac{1}{6}\left(mol\right)\)

\(\Rightarrow n_{ZnCl_2}=n_{Zn}+n_{ZnO}=\dfrac{4}{15}\left(mol\right)\) 

Mặt khác: \(m_{H_2}=0,1\cdot2=0,2\left(g\right)\)

\(\Rightarrow m_{dd}=m_{hh}+m_{ddHCl}-m_{H_2}=119,8\left(g\right)\) \(\Rightarrow C\%_{ZnCl_2}=\dfrac{\dfrac{4}{15}\cdot136}{119,8}\cdot100\%\approx30,27\%\)

c) Giả sử khí là SO₂

PTHH: \(Zn+H_2SO_{4\left(đ\right)}\xrightarrow[]{t^o}ZnSO_4+SO_2\uparrow+H_2O\)

Theo PTHH: \(n_{SO_2}=n_{Zn}=0,1\left(mol\right)\) \(\Rightarrow V_{SO_2}=0,1\cdot22,4=2,24\left(l\right)\)

Gọi \(\left\{{}\begin{matrix}n_{Ag}=a\left(mol\right)\\n_{FeO}=b\left(mol\right)\end{matrix}\right.\)

\(n_{SO_2}=\dfrac{1,344}{22,4}=0,6\left(mol\right)\)

PTHH:

\(2Ag+2H_2SO_4\rightarrow Ag_2SO_4+SO_2\uparrow+2H_2O\)

a             a                \(\dfrac{a}{2}\)             \(\dfrac{a}{2}\)

\(2FeO+4H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+SO_2\uparrow+4H_2O\)

b                 2b               \(\dfrac{b}{2}\)                \(\dfrac{b}{2}\)

Hệ pt

\(\left\{{}\begin{matrix}108a+72b=11,52\\\dfrac{a}{2}+\dfrac{b}{2}=0,06\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,08\left(mol\right)\\b=0,04\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}m_{Ag}=0,08.108=8,64\left(g\right)\\m_{FeO}=0,04.72=2,88\left(g\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\%m_{Ag}=\dfrac{8,64}{11,52}=75\%\\\%m_{FeO}=100\%-75\%=25\%\end{matrix}\right.\)

b, \(\rightarrow n_{H_2SO_4}=0,08+0,4.2=0,16\left(mol\right)\\ \rightarrow C_{MddH_2SO_4}=\dfrac{0,16}{0,8}=0,2M\)

c, \(n_{NaOH}=1,25.0,5=0,625\left(mol\right)\)

PTHH:

\(6NaOH+Fe_2\left(SO_4\right)_3\rightarrow2Fe\left(OH\right)_3+3Na_2SO_4\)

LTL: \(\dfrac{0,625}{6}>\dfrac{0,04}{2}\) => NaOH dư

Theo pthh:

\(\left\{{}\begin{matrix}n_{NaOH\left(pư\right)}=6n_{Fe_2\left(SO_4\right)_3}=6.0,04=0,24\left(mol\right)\\n_{Na_2SO_4}=3n_{Fe_2\left(SO_4\right)_3}=3.0,04=0,12\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}C_{MddNaOH\left(dư\right)}=\dfrac{0,24}{0,5}=0,48M\\C_{MddNa_2SO_4}=\dfrac{0,12}{0,5}=0,24M\end{matrix}\right.\)

\(n_{CO_2}=\dfrac{2,8}{22,4}=0,125\left(mol\right)\)

PTHH :

\(CO_2+2KOH\rightarrow K_2CO_3+H_2O\)

0,125    0,25          0,125 

\(a,C_{M\left(KOH\right)}=\dfrac{0,25}{0,1}=2,5\left(M\right)\)

\(b,C_{M\left(K_2CO_3\right)}=\dfrac{0,125}{0,1}=1,25\left(M\right)\)