2Al+3H2SO4->Al2(SO4)3+3H2

0,188------------------0,0944

n Al=0,188 mol

=>m Al2(SO4)3=0,0944.342=32,2848g

Đáp án+Giải thích các bước giải:

`n_{Al} = m/M = {5,4}/{27}` `= 0,2` `[mol]`

`PTHH :`

`2Al + 3H_2SO_4 -> Al_2[SO_4]_3 + 3H_2uparrow`

Dựa theo phương trình, ta có :

`n_{H_2} = 3/2n_{Al} = 3/2xx0,2 = 0,3` `[mol]`

`->` `V_{H_2 [đktc]} = nxx22,4 = 0,3xx22,4 = 6,72` `[l]`

2Fe+6H2SO4−−>Fe2(SO4)3+6H2O+3SO2

x--------------------------------------------------------------1,5x(mol)

Cu+2H2SO4−−>CuSO4+H2O+SO2

y---------------------------------------------------y(mol)

nSO2=5,622,4=0,25(mol)

Theo bài ta có hpt

{56x+64y=121,5

x+y=0,25

⇒{x=0,1y=0,1

mFe=0,1.56=5,6(g)

mCu=0,1.64=6,4(g)

=>%Fe=46,67%

=>%Cu=53,33%

=>VH2SO4=\(\dfrac{0,5}{1}\)=0,5l=500ml

\(\left(1\right)2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+8H_2O+5Cl_2\uparrow\\ \left(2\right)MnO_2+4HCl\rightarrow MnCl_2+2H_2O+Cl_2\uparrow\\ \left(3\right)2NaCl\underrightarrow{dpnc}2Na+Cl_2\uparrow\\\left(4\right)Cl_2+2NaOH\rightarrow NaCl+NaClO+H_2O\\ \left(5\right)CaOCl_2+2HCl\rightarrow CaCl_2+H_2O+Cl_2\uparrow\)

\(\left(6\right)2Fe+3Cl_2\underrightarrow{t^o}2FeCl_3\\ \left(7\right)Cl_2+2Na\underrightarrow{t^o}2NaCl\\ \left(8\right)Cl_2+H_2O⇌HCl+HClO\)

\(\left(9\right)NaCl+H_2SO_4\underrightarrow{t^o}NaHSO_4+HCl\\ \left(10\right)4HCl+MnO_2\rightarrow MnCl_2+2H_2O+Cl_2\uparrow\\ \left(11\right)Cl_2+H_2\underrightarrow{as}2HCl\\ \left(13\right)2HCl+Fe\rightarrow FeCl_2+H_2\uparrow\\ \left(14\right)2HCl+CuO\rightarrow CuCl_2+H_2\\ \left(15\right)FeCl_2+2AgNO_3\rightarrow2AgCl\downarrow+Fe\left(NO_3\right)_2\\ \left(16\right)CuCl_2+2AgNO_3\rightarrow2AgCl\downarrow+Cu\left(NO_3\right)_2\)

a) 

\(n_{Zn}=\dfrac{16,25}{65}=0,25\left(mol\right)\)

PTHH: Zn + H₂SO₄ --> ZnSO₄ + H₂

         0,25-->0,25------------->0,25

=> V_H2 = 0,25.22,4 = 5,6 (l)

b) \(C_{M\left(dd.H_2SO_4\right)}=\dfrac{0,25}{0,3}=\dfrac{5}{6}M\)

c) \(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂ --t^o--> 2Fe + 3H₂O

Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,25}{3}\) => Fe₂O₃ dư, H₂ hết

PTHH: Fe₂O₃ + 3H₂ --t^o--> 2Fe + 3H₂O

           \(\dfrac{0,25}{3}\) <--0,25----->\(\dfrac{0,5}{3}\)

=> \(m=32-\dfrac{0,25}{3}.160+\dfrac{0,5}{3}.56=28\left(g\right)\)

PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

đb:       0,25

a) số mol của Zn là: \(n_{Zn}=\dfrac{m_{Zn}}{M_{Zn}}=\dfrac{16,25}{65}=0,25\left(mol\right)\)

Theo PTHH, ta có: \(n_{H_2}=\dfrac{0,25\cdot1}{1}=0,25\left(mol\right)\)

Thể tích của H₂ ở đktc là: \(V_{H_2\left(đktc\right)}=n_{H_2}\cdot22,4=0,25\cdot22,4=5,6\left(l\right)\)

2 câu còn lại mk chịu

a) 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

b) \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

            0,2<----0,3<-----------------0,3

=> m_Al = 0,2.27 = 5,4 (g)

c) \(m_{dd.H_2SO_4}=\dfrac{0,3.98}{30\%}=98\left(g\right)\)

$a\big)$

$Zn+2HCl\to ZnCl_2+H_2$

$ZnCl_2:$ kẽm clorua

$b\big)$

$n_{Zn}=\dfrac{3}{65}(mol)$

Theo PT: $n_{H_2}=n_{Zn}=\dfrac{3}{65}(mol)$

$\to V_{H_2}=\dfrac{3}{65}.22,4\approx 1,034(l)$

$c\big)$

Theo PT: $n_{ZnCl_2}=n_{Zn}=\dfrac{3}{65}(mol)$

$\to m_{ZnCl_2}=\dfrac{3}{65}.136\approx 6,28(g)$

\(Cu\left(I,II\right)\\ Ag\left(I\right)\\ Zn\left(II\right)\\ Ba\left(II\right)\\ Hg\left(I,II\right)\\ Au\left(I,III\right)\\ Pb\left(II,IV\right)\)

Cu (II). Ag (I), Zn (II), Ba (II), Hg (II), Pb (II)

a)

\(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)

PTHH: 2Na + 2H₂O --> 2NaOH + H₂

              0,2-------------->0,2--->0,1

=> Chất tan trong dd X là NaOH

m_NaOH = 0,2.40 = 8 (g)

m_{dd sau pư} = 4,6 + 59,6 - 0,1.2 = 64 (g)

=> \(C\%=\dfrac{8}{64}.100\%=12,5\%\)

b) 

PTHH: CuO + H₂ --t^o--> Cu + H₂O

                       0,1------>0,1

=> m_Cu = 0,1.64 = 6,4 (g)

?