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

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

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

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

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

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

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

gọi R hóa trị 2

R+2H2SO4->RSO4+H2O+2SO2

ta có :\(\dfrac{6,4}{R}=0,2.\dfrac{1}{2}\)

=>R=64 ( đồng) 

=>R=Cu

nhúm quỳ 

-Chuyển đỏ :H2SO4

-Chuyển xanh :Ba(OH)2

- ko hiện tg:Nacl,NaI

ta sục khí Cl2

-Xuất hiện kết tủa đen :NaI

-Còn lại là NaCl

2NaI+Cl2->2NaCl+I2

b)

ta nhúm quỳ :

-Quỳ chuyển đỏ :HNO3

-Quỳ chuyển xanh :KOH

- quỳ ko chuyển màu KI, KBr

ta sục khí Cl2

-Xuất hiện kết tủa đen :KI

-Xuất hiện kết tảu vàng là KBr

2KI+Cl2->2KCl+I2

2KBr+Cl2->2KCl+Br2

\(a,n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\\ n_S=\dfrac{4,8}{32}=0,15\left(mol\right)\)

PTHH: Fe + S --t^o--> FeS

LTL: 0,4 > 0,15 => Fe dư

Y gồm: Fe dư và FeS

Theo pthh: \(n_{Fe\left(pư\right)}=n_{FeS}=n_S=0,15\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=\left(0,4-0,15\right).56=14\left(g\right)\\m_{FeS}=0,15.88=13,2\left(g\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{14}{14+13,2}.100\%=51,47\%\\\%m_{FeS}=100\%-51,47\%=48,53\%\end{matrix}\right.\)

b) PTHH:

Fe + 2HCl ---> FeCl₂ + H₂

0,25->0,5---------------->0,25

FeS + 2HCl ---> FeCl₂ + H₂S

0,15->0,3---------------->0,15

\(m_{HCl}=\left(0,5+0,3\right).36,5=29,2\left(g\right)\\ m=\dfrac{29,2}{3,65\%}=800\left(g\right)\)

\(M_Z=\dfrac{0,15.34+0,25.2}{0,15+0,25}=14\left(\dfrac{g}{mol}\right)\\ M_X=\dfrac{0,1.32+0,3.28}{0,1+0,3}=29\left(\dfrac{g}{mol}\right)\)

=> d_Z/X = \(\dfrac{14}{29}\approx0,48\)