Gọi hóa trị với H là a => Hóa trị cao nhất là 3a

=> a + 3a = 8 => a = 2

Ta có: \(\%X=\dfrac{M_X}{M_X+48}.100\%=40\%\)

=> \(M_X=32\left(g/mol\right)\)

=> X là lưu huỳnh

chụp thẳng đc ko chứ mỏi cổ quá

a) \(n_{H_2}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\)

PTHH: A + 2HCl ---> ACl₂ + H₂ 

          0,2<-0,4<-------0,2<---0,2

=> \(M_A=\dfrac{11,2}{0,2}=56\left(g/mol\right)\)

=> A là sắt (Fe)

b) \(C\%_{HCl}=\dfrac{0,4.36,5}{200}.100\%=7,3\%\)

c) \(m_{dd}=11,2+200-0,2.2=210,8\left(g\right)\)

=> \(C\%_{FeCl_2}=\dfrac{0,2.127}{210,8}.100\%=12,05\%\)

+ CuCl₂ có M = 64 + 35,5.2 = 135 (g/mol)

+ Zn(NO₃)₂ có M = 65 + (14 + 16.3).2 = 189 (g/mol)

a) \(n_{AgNO_3}=0,01.1=0,01\left(mol\right)\)

PTHH: 2AgNO₃ + Cu ---> Cu(NO₃)₂ + 2Ag

             0,01----->0,005--->0,005----->0,01

            Mg + 2HCl ---> MgCl₂ + H₂

=> m_A = 0,01.108 = 1,08 (g)

b) Mg(NO₃)₂ + 2NaOH ---> Mg(OH)₂ + 2NaNO₃ 

      0,005------->0,01

=> \(V_{ddNaOH}=\dfrac{0,01}{1}=0,01\left(l\right)=10\left(ml\right)\)

a) \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right);n_{H_2SO_4}=0,2.2,5=0,5\left(mol\right)\)

PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

Xét tỉ lệ: \(\dfrac{0,3}{2}< \dfrac{0,5}{3}\) => H₂SO₄ dư, Al hết, tính theo Al

Theo PTHH: \(n_{H_2SO_4\left(pư\right)}=n_{H_2}=\dfrac{3}{2}n_{Al}=0,45\left(mol\right)\)

=> \(V_A=V_{H_2}=0,45.24,79=11,1555\left(l\right)\)

b) Theo PTHH: \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,15\left(mol\right)\)

=> \(m_{muối}=m_{Al_2\left(SO_4\right)_3}=0,15.342=51,3\left(g\right)\)

\(n_{H_2SO_4\left(dư\right)}=0,5-0,45=0,05\left(mol\right)\)

=> \(\left\{{}\begin{matrix}C_{M\left(Al_2\left(SO_4\right)_3\right)}=\dfrac{0,15}{0,2}=0,75M\\C_{M\left(H_2SO_4.dư\right)}=\dfrac{0,05}{0,2}=0,25M\end{matrix}\right.\)

0,56 gam chất rắn không tan là Fe => \(n_{Fe}=\dfrac{0,56}{56}=0,01\left(mol\right)\)

1,6 gam chất rắn màu đỏ là Cu => \(n_{Cu}=\dfrac{1,6}{64}=0,025\left(mol\right)\)

PTHH:
2Al + 3CuSO₄ ---> Al₂(SO₄)₃ + 3Cu (1)

Fe + CuSO₄ ---> FeSO₄ + Cu (2)

2NaOH + 2Al + 2H₂O ---> 2NaAlO₂ + 3H₂ (3)
Theo PT (1), (2): \(n_{Cu}=\dfrac{3}{2}n_{Al}+n_{Fe}\)

=> \(n_{Al}=\dfrac{2}{3}.\left(0,025-0,01\right)=0,01\left(mol\right)\)

=> a = 0,01.27 + 0,01.56 = 0,83 (g)

PTHH:     \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\) (1)

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

\(n_{CuO}=\dfrac{4,8}{80}=0,06\left(mol\right)\)

Theo PT (1), (2): \(n_{Al}=\dfrac{2}{3}.n_{H_2}=\dfrac{2}{3}.n_{CuO}=\dfrac{2}{3}.0,06=0,04\left(mol\right)\)

=> m_Al = 0,04.27 = 1,08 (g)

Gọi \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\)

=> 27a + 56b = 22,2 (1)
\(n_{H_2}=\dfrac{14,874}{24,79}=0,6\left(mol\right)\)

PTHH:

2Al + 6HCl ---> 2AlCl₃ + 3H₂ 

a--------------------------->1,5a

Fe + 2HCl ---> FeCl₂ + H₂ 

b-------------------------->b

=> 1,5a + b = 0,6 (2)
Từ (1), (2) => a = 0,2; b = 0,3

=> \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2.27}{22,2}.100\%=24,32\%\\\%m_{Fe}=100\%-24,32\%=75,68\%\end{matrix}\right.\)

a) \(n_{H_2}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\)

Chất rắn không tan là Cu => m_Cu = 5,25 (g)

PTHH: Zn + H₂SO₄ ---> ZnSO₄ + H₂

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

=> m_Zn = 0,15.65 = 9,75 (g)

b) \(m_{ddH_2SO_4}=\dfrac{0,15.98}{24,5\%}=60\left(g\right)\)