\(PTK_A=23.2=46\left(đvC\right)\)

\(n_{Ba\left(OH\right)_2}=\dfrac{100.3,42\%}{171}=0,02\left(mol\right)\)

\(n_{HCl}=\dfrac{100.1,095\%}{36,5}=0,03\left(mol\right)\)

PTHH: \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

Xét tỉ lệ: \(\dfrac{0,02}{1}>\dfrac{0,03}{2}\) => Ba(OH)₂ dư, HCl hết

PTHH: \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

               0,015<-----0,03---->0,015

\(\left\{{}\begin{matrix}C\%_{BaCl_2}=\dfrac{0,015.208}{100+100}.100\%=1,56\%\\C\%_{Ba\left(OH\right)_2\left(dư\right)}=\dfrac{\left(0,02-0,015\right).171}{100+100}.100\%=0,4275\%\end{matrix}\right.\)

\(n_{NO}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Gọi số mol Al, Fe là a, b (mol)

=> 27a + 56b = 11 (1)

\(Al+4HNO_3\rightarrow Al\left(NO_3\right)_3+NO+2H_2O\)

  a------>4a----------->a------->a

\(Fe+4HNO_3\rightarrow Fe\left(NO_3\right)_3+NO+2H_2O\)

  b------>4b----------->b-------->b

=> a + b = 0,3 (2)

(1)(2) => a = 0,2 (mol); b = 0,1 (mol)

\(\left\{{}\begin{matrix}m_{Al}=0,2.27=5,4\left(g\right)\\m_{Fe}=0,1.56=5,6\left(g\right)\end{matrix}\right.\)

\(\left\{{}\begin{matrix}m_{Al\left(NO_3\right)_3}=0,2.213=42,6\left(g\right)\\m_{Fe\left(NO_3\right)_3}=0,1.242=24,2\left(g\right)\end{matrix}\right.\) => m_muối = 42,6 + 24,2 = 66,8 (g)

\(m_{HNO_3}=\left(4a+4b\right).63=75,6\left(g\right)\)

Giả sử có 100 (g) dd HCl 32,85%

\(n_{HCl\left(bđ\right)}=\dfrac{100.32,85\%}{36,5}=0,9\left(mol\right)\)

Gọi số mol CaCO₃ là a (mol)

PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

                 a------->2a-------->a----->a

=> m_{dd X} = 100a + 100 - 44a = 100 + 56a (g)

n_HCl(X) = 0,9 - 2a (mol)

\(C\%_{HCl\left(X\right)}=\dfrac{36,5\left(0,9-2a\right)}{100+56a}.100\%=24,195\%\)

=> a = 0,1 (mol)

m_{dd X} = 105,6 (g)

n_HCl(X) = 0,7 (mol)

Gọi số mol MgCO₃ là b (mol)

PTHH: \(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\)

                  b-------->2b------->b------->b

m_{dd Y} = 105,6 + 84b - 44b = 105,6 + 40b (g)

n_HCl(Y) = 0,7 - 2b (mol)

\(C\%_{HCl\left(Y\right)}=\dfrac{\left(0,7-2b\right).36,5}{105,6+40b}.100\%=21,11\%\)

=> b = 0,04 (mol)

dd Y chứa \(\left\{{}\begin{matrix}CaCl_2:0,1\left(mol\right)\\MgCl_2:0,04\left(mol\right)\end{matrix}\right.\)

m_{dd Y} = 107,2 (g)

\(\left\{{}\begin{matrix}C\%_{CaCl_2}=\dfrac{0,1.111}{107,2}.100\%=10,35\%\\C\%_{MgCl_2}=\dfrac{0,04.95}{107,2}.100\%=3,54\%\end{matrix}\right.\)

Gọi \(\left\{{}\begin{matrix}C_{M\left(A\right)}=aM\\C_{M\left(B\right)}=bM\end{matrix}\right.\)

TN1:

\(\left\{{}\begin{matrix}n_{H_2SO_4\left(bd\right)}=0,05a\left(mol\right)\\n_{NaOH\left(bd\right)}=0,05b\left(mol\right)\end{matrix}\right.\)

n_NaOH(thêm) = 0,1.0,02 = 0,002 (mol)

PTHH: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

              0,1a<----0,05a

=> 0,1a = 0,05b + 0,002 (1)

- Xét TN2:

\(\left\{{}\begin{matrix}n_{H_2SO_4\left(bd\right)}=0,05a\left(mol\right)\\n_{NaOH\left(bd\right)}=0,1b\left(mol\right)\end{matrix}\right.\)

n_HCl(thêm) = 0,02.0,1 = 0,002 (mol)

PTHH: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

                0,1a<---0,05a

             \(NaOH+HCl\rightarrow NaCl+H_2O\)

               0,002<--0,002

=> 0,1a + 0,002 = 0,1b (2)

(1)(2) => a = 0,0; b = 0,08

a) Xét \(\dfrac{PTK_{CaO}}{PTK_{H_2O}}=\dfrac{56}{18}=3,11\)

=> Phân tử CaO nặng hơn phân tử H₂O 3,11 lần

Xét \(\dfrac{PTK_{CaO}}{PTK_{BaCO_3}}=\dfrac{56}{197}=0,284\)

=> Phân tử CaO nhẹ hơn phân tử BaCO₃ và bằng 0,284 lần

b) Xét \(\dfrac{5.PTK_{CO_2}}{PTK_{H_2}}=\dfrac{5.44}{2}=110\)

=> 5 phân tử CO₂ nặng hơn phân tử H₂ 110 lần

c) \(PTK_{X_3}=2.24=48\left(đvC\right)\)

=> NTK_X = 16 (đvC)

=> X là Oxi

\(BaO+H_2O\rightarrow Ba\left(OH\right)_2\)

\(Ba\left(OH\right)_2+Al_2O_3\rightarrow Ba\left(AlO_2\right)_2+H_2O\)

\(Ba\left(AlO_2\right)_2+2CO_2+4H_2O\rightarrow Ba\left(HCO_3\right)_2+2Al\left(OH\right)_3\downarrow\)

\(FeO+CO\underrightarrow{t^o}Fe+CO_2\)

\(2NaOH+Al_2O_3\rightarrow2NaAlO_2+H_2O\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

\(FeSO_4+2NaOH\rightarrow Fe\left(OH\right)_2\downarrow+Na_2SO_4\)

\(4Fe\left(OH\right)_2+O_2\underrightarrow{t^o}2Fe_2O_3+4H_2O\)

đề hỏi gì vậy bn 

1) \(2Li+2H_2O\rightarrow2LiOH+H_2\)

2) \(2K+2H_2O\rightarrow2KOH+H_2\)

3) \(Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\)

4) Không pư

5) \(Ba+2H_2O\rightarrow Ba\left(OH\right)_2+H_2\)

6) \(2Na+2H_2O\rightarrow2NaOH+H_2\)

7) Không pư

8) \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

9) \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

10) \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

11) \(2Zn+O_2\underrightarrow{t^o}2ZnO\)

12) Không pư

13) Không pư

14) \(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

15) Không pư

16) \(K_2O+H_2O\rightarrow2KOH\)

17) \(BaO+H_2O\rightarrow Ba\left(OH\right)_2\)

18) Không pư

19) \(CO_2+H_2O⇌H_2CO_3\)

20) \(P_2O_5+3H_2O\rightarrow2H_3PO_4\)

21) \(N_2O_5+H_2O\rightarrow HNO_3\)

22) \(SO_2+H_2O⇌H_2SO_3\)

23) \(SO_3+H_2O\rightarrow H_2SO_4\)

24) \(Cl_2+H_2O⇌HCl+HClO\)

25) \(Mg+2HCl\rightarrow MgCl_2+H_2\)

26) \(Fe+2HCl\rightarrow FeCl_2+H_2\)

27) \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

28) Không pư