a, \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

\(FeCl_3+3KOH\rightarrow3KCl+Fe\left(OH\right)_{3\downarrow}\)

\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

Theo PT: \(n_{HCl}=6n_{Fe_2O_3}=0,6\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,6.36,5=21,9\left(g\right)\)

b, \(n_{Fe\left(OH\right)_3}=n_{FeCl_3}=2n_{Fe_2O_3}=0,2\left(mol\right)\)

\(\Rightarrow m_{Fe\left(OH\right)_3}=0,2.107=21,4\left(g\right)\)

\(n_{KOH}=3n_{FeCl_3}=0,6\left(mol\right)\)

\(\Rightarrow C_{M_{KOH}}=\dfrac{0,6}{0,2}=3\left(M\right)\)

\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

PT: \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

a, \(n_{HCl}=6n_{Fe_2O_3}=0,6\left(mol\right)\)

\(\Rightarrow C\%_{HCl}=\dfrac{0,6.36,5}{500}.100\%=4,38\%\)

b, \(n_{FeCl_3}=2n_{Fe_2O_3}=0,2\left(mol\right)\)

PT: \(FeCl_3+3KOH\rightarrow3KCl+Fe\left(OH\right)_{3\downarrow}\)

______0,2_______0,6______________0,2 (mol)

\(\Rightarrow C_{M_{KOH}}=\dfrac{0,6}{0,2}=3\left(M\right)\)

\(m_{Fe\left(OH\right)_3}=0,2.107=21,4\left(g\right)\)

\(a)n_{MnO_2}=\dfrac{69,6}{87}=0,8mol\\ MnO_2+4HCl\xrightarrow[nhẹ]{đun}MnCl_2+Cl_2+H_2O\)

0,8            3,2               0,8            0,8             0,8

\(V_A=V_{Cl_2}=0,8.22,4=17,92l\\ b)Cl_2+2NaOH\rightarrow NaCl+NaClO+H_2O\)

    0,8       1,6               0,8         0,8

\(V_{ddNaOH}=\dfrac{1,6}{1}=1,6l\\ C_{M_{NaCl}}=\dfrac{0,8}{1,6}=0,5M\\ C_{M_{NaClO}}=\dfrac{0,8}{1,6}=0,5M\)

n_Mg = 0,1(mol)

PTHH: Mg + 2HCl --> MgCl₂ +H₂

n_Mg = n_MgCl2= n_H2 = 0,1(mol)

=> m_muối = 9,5(g) 

_VH2 = 2,24(l)

b) CM_HCl = 0,2/0,1=2(M)

PTHH: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

            \(CuCl_2+2KOH\rightarrow2KCl+Cu\left(OH\right)_2\downarrow\)

a+b) Ta có: \(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)

\(\Rightarrow n_{HCl}=0,2\left(mol\right)=n_{KOH}\) \(\Rightarrow\left\{{}\begin{matrix}C\%_{HCl}=\dfrac{0,2\cdot36,5}{300}\cdot100\%\approx2,43\%\\C_{M_{KOH}}=\dfrac{0,2}{0,2}=1\left(M\right)\end{matrix}\right.\)

c) PTHH: \(Cu\left(OH\right)_2\xrightarrow[]{t^o}CuO+H_2O\)

Theo các PTHH: \(n_{CuO\left(lý.thuyết\right)}=n_{Cu\left(OH\right)_2}=n_{Cu}=0,1\left(mol\right)\)

\(\Rightarrow n_{CuO}=0,1\cdot95\%=0,095\left(mol\right)\) \(\Rightarrow m_{CuO}=0,095\cdot80=7,6\left(g\right)\)

\(a,n_{H_2SO_4}=0,3.0,75+0,3.0,25=0,3\left(mol\right)\\ V_{ddH_2SO_4}=300+300=600\left(ml\right)=0,6\left(l\right)\\ \rightarrow C_{M\left(H_2SO_4\right)}=\dfrac{0,3}{0,6}=0,5M\\ m_{H_2SO_4}=0,3.98=29,4\left(g\right)\\ m_{ddH_2SO_4}=600.1,02=612\left(g\right)\\ \rightarrow C\%_{H_2SO_4}=\dfrac{29,4}{612}.100\%=4,8\%\)

\(b,\) Đặt kim loại M có hoá trị n (n ∈ N*)

PTHH: \(2M+nH_2SO_4\rightarrow M_2\left(SO_4\right)_n+nH_2\uparrow\)

           \(\dfrac{0,6}{n}\)<---0,3--------------------------->0,3

\(\rightarrow M_M=\dfrac{5,4}{\dfrac{0,6}{n}}=9n\left(g\text{/}mol\right)\)

Vì n là hoá trị của M nên ta xét bảng

Vậy M là Al

\(c,n_{KClO_3}=\dfrac{15,3125}{122,5}=0,125\left(mol\right)\)

PTHH:

\(2H_2+O_2\xrightarrow[]{t^o}2H_2O\)

0,3-->0,15

\(2KClO_3\xrightarrow[]{t^o}2KCl+3O_2\uparrow\)

0,1<---------------------0,15

\(\rightarrow H=\dfrac{0,1}{0,125}.100\%=80\%\)

a)

$MnO_2 + 4HCl \to MnCl_2 + Cl_2 + 2H_2O$

Theo PTHH :

n Cl2 = n MnO2 = 10,44/87 = 0,12(mol)

=> V Cl2 = 0,12.22,4 = 2,688(lít)

b)

$2NaOH + Cl_2 \to NaCl + NaClO + H_2O$

n NaOH = 2n Cl2 = 0,24(mol)

=> V dd NaOH = 0,24/2 = 0,12(lít)

PTHH: \(CaCl_2+Na_2CO_3\rightarrow2NaCl+CaCO_3\downarrow\)

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

A là dd NaCl

B là CaCO₃

C là dd NaNO₃

D là AgCl

Ta có: \(n_{AgCl}=\dfrac{57,4}{143,5}=0,4\left(mol\right)=n_{NaCl}\)

\(\Rightarrow n_{CaCO_3}=0,2\left(mol\right)=n_{Na_2CO_3}\) \(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=0,4\cdot58,5=23,4\left(g\right)\\m_{CaCO_3}=0,2\cdot100=20\left(g\right)\\m_{ddNa_2CO_3}=\dfrac{0,2\cdot106}{16\%}=132,5\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{ddA}=m_{ddCaCl_2}+m_{ddNa_2CO_3}-m_{CaCO_3}=290,1\left(g\right)\)

\(\Rightarrow C\%_{NaCl}=\dfrac{23,4}{290,1}\cdot100\%\approx8,07\%\)