\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

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

=> m_Fe = 0,2.56 = 11,2 (g)

\(n_{CuS}=\dfrac{9,6}{96}=0,1\left(mol\right)\)

PTHH: Cu(NO₃)₂ + H₂S --> CuS + 2HNO₃

                                 0,1<---0,1

             FeS + 2HCl --> FeCl₂ + H₂S

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

=> m_FeS = 0,1.88 = 8,8 (g)

=> m = 11,2 + 8,8 = 20 (g)

Gọi số mol O₂, O₃ là a, b (mol)

Có: \(\dfrac{32a+48b}{a+b}=16,2.2=32,4\left(g/mol\right)\)

=> 0,4a = 15,6b

=> a = 39b

\(\%V_{O_3}=\dfrac{b}{a+b}.100\%=\dfrac{b}{39b+b}.100\%=2,5\%\)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: Fe + H₂SO₄ --> FeSO₄ + H₂

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

=> \(n_{Ag}=\dfrac{22-0,2.56}{108}=0,1\left(mol\right)\)

=> \(\%n_{Ag}=\dfrac{0,1}{0,1+0,2}.100\%=33,33\%\)

Bảo toàn Fe: n_Fe(bđ) = 0,012 + 0,12 + 0,01.3 + 0,01.2 = 0,182 (mol)

=> m = 0,182.56 = 10,192 (g)

Gọi \(n_{O_3\left(pư\right)}=a\left(mol\right)\left(đk:a>0\right)\)

PTHH: \(2Ag+O_3\rightarrow Ag_2O+O_2\uparrow\)

                      a---------------->a

\(m_{tăng}=m_{O_3\left(pư\right)}-m_{O_2\left(sinh.ra\right)}=48a-32a=1,28\\ \Leftrightarrow a=0,08\left(mol\right)\left(TM\right)\\ n_X=\dfrac{44,8}{22,4}=2\left(mol\right)\\ \rightarrow\%V_{O_3}=\dfrac{0,08}{2}.100\%=8\%\)

hỏi 1 lần thôi bn :D

AXIT: HCl ( axit clohidric)  , HNO3 ( axit nitric)

BAZO: NaOH ( Natri hidroxit)

MUỐI: FeSO4 ( Sắt Sunfat)  , KCl (Kali clorua), Al2(SO4)3 ( Nhôm Sunfat)

ÕXIT ÃXIT: SO3 ( Lưu huỳnh trioxit) , P2O5 ( ddiPhotpho pentaoxit)

ÕXIT BAZO: FeO (Sắt (2) Oxit), MgO ( Magie Oxit)

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

hỏi 1 lần thôi bn :v

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

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

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