\(n_{Al}=\dfrac{10.8}{27}=0.4\left(mol\right)\)

\(n_{O_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)

\(4Al+3O_2\underrightarrow{t^0}2Al_2O_3\)

\(0.4......0.3..........0.2\)

\(Al_2O_3:\) oxit bazo => Nhôm oxit

\(m_{Al_2O_3}=0.2\cdot102=20.4\left(g\right)\)

\(\rightarrow m_{Al_2O_3}=0,4.102=40,8\left(g\right)\)\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)

\(n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

\(4Al+3O_2\rightarrow2Al_2O_3\)

-> Phân loại: Phản ứng hóa hợp

-> \(Al_2O_3\) : nhôm oxit 

b) So sánh:

\(\dfrac{n_{Al}}{4}=0,1< \dfrac{n_{O_2}}{3}=0,1333\)

=> Al hết 

\(n_{Al_2O_3}=\dfrac{1}{2}.n_{Al}=0,2\left(mol\right)\)

\(\text{mAgNO3=85.16%=13,6g}\)

=> nAgNO3=0,08mol.

Ta có:

\(\text{nHCl=8,96:22,4=0,4mol}\)

=> mHCl=14,6g

=> C% A= 14,6.100:(184,5+14,6)= 7,332%.

Vậy 50g A có 50.7,332%=3,666g HCl => nHCl=0,1mol.

Ta có PTHH

\(\text{AgNO3+HCl=AgCl+HNO3}\)

=> Sau thu đc 0,08 kết tủa AgCl, 0,08g HNO3 và dư 0,02 mol HCl dư.

\(\text{Khối lượng kết tủa: 0,08.143,5=11,48g. }\)

Nồng độ % các chất tan trong B:

\(\text{C%HNO3=(0,08.63.100)/(50+85-11,48)=4,08%;}\)

\(\text{C%HCl dư=(0,02.36,5.100)/(50+85-11,48)=0,59%}\)

\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+3H_2\\ Đặt:n_{Al}=a\left(mol\right);n_{Fe}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}27a+56b=11\\1,5a+b=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\\ a,n_{HCl}=2.n_{H_2}=2.0,4=0,8\left(mol\right)\\ \Rightarrow V_{ddHCl}=\dfrac{0,8}{8}=0,1\left(l\right)\\ b,FeCl_2+2AgNO_3\rightarrow Fe\left(NO_3\right)_2+2AgCl\downarrow\\ AlCl_3+3AgNO_3\rightarrow Al\left(NO_3\right)_3+3AgCl\downarrow\\ n_{AgCl}=n_{AgNO_3}=3.n_{AlCl_3}+2.n_{FeCl_2}=3.a+2.b=3.0,2+2.0,1=0,8\left(mol\right)\\ \Rightarrow a=\dfrac{170.0,8}{250}.100=54,4\%\\ b=m_{\downarrow}=m_{AgCl}=0,8.143,5=114,8\left(g\right)\)

AlCl3 + 3AgNO3 -> 3AgCl + Al(NO3)3 (1)

nAlcl3=m/M= 13.35/133.5=0.1 mol

theo pt

nAgcl= nAgno3=3nAlcl3=3.0.1=0.3 mol; nAl(no3)3=nAlcl3=0.1 mol

100ml = 0.1 lít

=> CM Agno3 = n/V =0.3/0.1=3 M

mAgcl=n.M=0.3.143.5=43.05 g

mAl(no3)3=n.M=0.1.213=21.3 g

n_AlCl3=\(\dfrac{13,35}{133,5}=0,1\left(mol\right)\)

pthh:

AlCl₃ + 3AgNO₃ \(\rightarrow\) Al(NO3)₃ + 3AgCl

0,1... .... ....0,3... ... ... ...0,1... ... ... ...0,3 (mol)

a, m_AgCl=n.M=0,3.143,5=43,05(g)

b, CM_AgNO3=\(\dfrac{n}{V}=\dfrac{0,3}{0,1}=3\left(M\right)\)

c, m_{muối Al(NO3)3}=n.M=0,1.213=21,3(g)

* Tính CO2 và H2O nhé:

Ca(OH)2 + CO2 ---> CaCO3 + H2O

0,09 ----->0,09--------->0,09

CO2 + CaCO3 + H2O ----> Ca(HCO3)2

x---------->x------------------------------->x

Ca(HCO3)2 + Ba(OH)2 -----> CaCO3 + BaCO3 + 2H2O

x--------------------->x------------------->x----------->x

-> 18,85 = 100(0,09 - x) + 100x + 197x

--> x = 0,05 mol

--> nCO2 = 0,05 + 0,09 = 0,14 mol

nH2O = (mktua + mdd tăng - mCO2)/18 = 0,09 mol

nhận thấy số mol H2O < n CO2 ---> hợp chất này có công thức tổng quát là CnH2n-3NH2 hay CnH2n-5N

CnH2n-5N + (3n-5/2) O2 -----> nCO2 + (n-5/2) H2O + 1/2 N2

-----a--------------------------------------na-------(n-5/2)a

--> 5a/2 = nCO2 - nH2O = 0,05 mol

--> a = 0,02 mol

M(A) = 2,14/0,02 = 107

--> n = 7

--> C7H9N

a, Ba(NO3)2+ Na2CO3= BaCO3+ 2NaNO3

b,

nBaCO3= 4,925/197= 0,025 mol= nBa(NO3)2

=> mBa(NO3)2= 0,025.261= 6,525g

=> mNaCl= 8,5-6,525= 1,975g

%Ba(NO3)2= 6,525.100:8,5= 76,76%

%NaCl= 23,24%