\(m_{dd.H_2SO_4}=51,5.1,84=94,76\left(g\right)\)

\(n_{SO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: 2Al + 6H₂SO₄ --> Al₂(SO₄)₃ + 3SO₂ + 6H₂O

            0,2<----0,6<--------------------0,3

=> m_Al = 0,2.27 = 5,4 (g)

=> \(n_{Fe_2O_3}=\dfrac{21,4-5,4}{160}=0,1\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂SO₄ --> Fe₂(SO₄)₃ + 3H₂O

                0,1--->0,3

=> \(m_{H_2SO_4\left(lý.thuyết\right)}=\left(0,6+0,3\right).98=88,2\left(g\right)\)

=> \(m_{H_2SO_4\left(tt\right)}=88,2.105\%=92,61\left(g\right)\)

=> \(C\%=\dfrac{92,61}{94,76}.100\%=97,7\%\)

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

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

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

\(n_{Fe}=\dfrac{m}{M}=\dfrac{2,8}{56}=0,05\left(mol\right)\)

0,05                                     →           0,075                         ( mol )

\(V_{SO_2}=n.22,4=0,075.22,4=1,68\left(l\right)\)

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

a)2Al+6HCl→2AlCl3+3H2

Fe+2HCl→FeCl2+H2

2Al+6H2SO4→Al2(SO4)3+3SO2+6H2O

2Fe+6H2SO4→Fe2(SO4)3+3SO2+6H2O

Cu+2H2SO4→CuSO4+SO2+2H2O

hh:Al(amol),Fe(bmol),Cu(cmol)

nNaOH=0,2×2=0,4mol

nHCl=0,4×2=0,8mol

⇒nHClpu=0,8−0,4=0,4mol

nSO2=5,6\22,4=0,25mol

27a+56b+64c=14,2

0,5a×3+0,5b×2=0,4

0,5a×1,5+0,5b×1,5+0,5c=0,25

⇒a=0,2;b=0,1;c=0,05

mAl=0,2×27=5,4g

mFe=0,1×56=5,6g

mCu=0,05×64=3,2g

b)mddspu=7,1+50−0,25×64=41,1g

C%Al2(SO4)3=41,6%

C%Fe2(SO4)3=24,33%

C%CuSO4=9,73%

chị ơi cho em hỏi cái 0,5 đứng trước a khi lập hệ là sao ạ?

hỏi 1 lần thôi bn :v