a. \(n_{Na_2O}=\dfrac{6.2}{62}=0,1\left(mol\right)\)

PTHH : Na₂O + H₂O -> 2NaOH 

               0,1                     0,2

b. \(C_{M_{dd}NaOH}=\dfrac{0,2}{0,1}=2M\)

c. PTHH : 2NaOH + H₂SO₄ -> Na₂SO₄ + 2H₂O 

                     0,2           0,1             0,1

\(V_{dd_{H_2SO_4}}=\dfrac{0,1}{2}=0,05\left(l\right)\)

\(m_{Na_2SO_4}=0,1.142=14,2\left(g\right)\)

a) 8,3 gam M có \(\left\{{}\begin{matrix}n_{C_2H_4}=a\left(mol\right)\\n_{C_3H_6}=b\left(mol\right)\\n_{C_2H_2}=c\left(mol\right)\end{matrix}\right.\left(a,b,c>0\right)\)

`=>` \(28a+42b+26c=8,3\left(1\right)\)

\(n_{Br_2}=\dfrac{48}{160}=0,3\left(mol\right)\)

PTHH:

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

a-------->a

\(C_3H_6+Br_2\rightarrow C_3H_6Br_2\)

b-------->b

\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

c-------->2c

`=>` \(a+b+2c=0,3\left(2\right)\)

8,4l hh M có \(\left\{{}\begin{matrix}n_{C_2H_4}=ak\left(mol\right)\\n_{C_3H_6}=bk\left(mol\right)\\n_{C_2H_2}=ck\left(mol\right)\end{matrix}\right.\left(k>0\right)\)

\(n_M=\dfrac{8,4}{22,4}=0,375\left(mol\right)\) `=>` \(ak+bk+ck=0,375\left(3\right)\)

\(n_{C_2Ag_2}=\dfrac{18}{240}=0,075\left(mol\right)\)

PTHH: \(C_2H_2+2AgNO_3+2NH_3\rightarrow2NH_4NO_3+C_2Ag_2\downarrow\)

            0,075<---------------------------------------------0,075

`=>` \(ck=0,075\left(4\right)\)

Từ \(\left(1\right),\left(2\right),\left(3\right),\left(4\right)\Rightarrow\left\{{}\begin{matrix}a=b=0,1\\c=0,05\\k=\dfrac{3}{2}\end{matrix}\right.\)

`=>` \(\left\{{}\begin{matrix}\%V_{C_2H_4}=\%V_{C_3H_6}=\dfrac{0,1}{0,1+0,1+0,05}.100\%=40\%\\\%V_{C_2H_2}=100\%-40\%-40\%=20\%\end{matrix}\right.\)

b) 8,4l M có \(\left\{{}\begin{matrix}n_{C_2H_4}=n_{C_3H_6}=\dfrac{3}{2}.0,1=0,15\left(mol\right)\\n_{C_2H_2}=\dfrac{3}{2}.0,05=0,075\left(mol\right)\end{matrix}\right.\)

Quy đổi H₂ và Br₂ thành X₂; \(n_{Br_2}=\dfrac{40}{160}=0,25\left(mol\right)\)

PTHH:

\(C_2H_4+X_2\rightarrow C_2H_4X_2\)

0,15--->0,15

\(C_3H_6+X_2\rightarrow C_3H_6X_2\)

0,15--->0,15

\(C_2H_2+2X_2\rightarrow C_2H_2X_4\)

0,075->0,15

`=>` \(\sum n_{X_2}=0,15+0,15+0,15=0,45\left(mol\right)\)

Ta có: \(n_{X_2}=n_{H_2}+n_{Br_2}\)

`=>` \(a=n_{H_2}=0,45-0,25=0,2\)

a, PT: \(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

b, Ta có: \(n_{AlCl_3}=\dfrac{26,7}{133,5}=0,2\left(mol\right)\)

Theo PT: \(n_{Al_2O_3}=\dfrac{1}{2}n_{AlCl_3}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al_2O_3}=\dfrac{0,1.102}{15}.100\%=68\%\\\%m_{Cu}=32\%\end{matrix}\right.\)

CuCl₂ + 2NaOH--> 2NaCl + Cu(OH)₂
3Fe(NO3)₃ +6 NaOH--> 6NaNO₃ +3 Fe(OH)2
Fe(SO₄)₃+3BaCl2 ---> 2FeCl₃ + 3BaSO₄
FeCl₃ +3 NaNO₃ --> Fe(NO₃)₃ + 3NaCl
Na₂CO₃ +Ba(OH)₂ --> BaCO₃ + 2NaOH
BaCO₃ + H₂SO₄ --> BaSO₄ +H₂O+CO₂

giúp em với ạ

a)

\(n_{Ba\left(OH\right)_2}=0,8.1=0,8\left(mol\right)\)

\(n_{BaCO_3}=\dfrac{118,2}{197}=0,6\left(mol\right)\)

PTHH: \(Ba\left(OH\right)_2+CO_2\rightarrow BaCO_3+H_2O\)

            \(BaCO_3+CO_2+H_2O\rightarrow Ba\left(HCO_3\right)_2\)

Gọi số mol CO₂ là a (mol); số mol H₂O là b (mol)

Ta có: \(m_{giảm}=m_{BaCO_3}-m_{CO_2}-m_{H_2O}\)

=> \(118,2-44a-18b=68,8\left(g\right)\)

=> 44a + 18b = 49,4 

TH1: Nếu kết tủa không bị hòa tan

=> a = 0,6 (mol)

=> b = \(\dfrac{23}{18}\) (mol)

\(n_X=\dfrac{5,6}{22,4}=0,25\left(mol\right)\Rightarrow\overline{C}=\dfrac{0,6}{0,25}=2,4\left(L\right)\)

TH2: Nếu kết tủa bị hòa tan 1 phần

Bảo toàn Ba: \(n_{Ba\left(HCO_3\right)_2}=0,2\left(mol\right)\)

Bảo toàn C: a = 1 (mol)

=> b = 0,3 (mol)

\(\overline{C}=\dfrac{1}{0,25}=4\)

=> A, B có CTPT lần lượt là C₄H_n (x mol) và C₄H_2+n (y mol)

Có: x + y = 0,25

Bảo toàn H: xn + yn + 2y = 0,6

=> 0,25n + 2y = 0,6

Mà 0 < y < 0,25

=> n < 2,4

=> n = 2 (do n luôn chẵn)

Vậy A là C₄H₂, B là C₄H₄

Thay số => x = 0,2 ; y = 0,05 

=> n_A : n_B = 4 : 1

CTCT:

- Của A:

(1) \(CH\equiv C-C\equiv CH\)

- Của B:

(1) \(CH\equiv C-CH=CH_2\)

b)

Gọi số mol C₄H₂, C₄H₄ là u, v (mol)

=> \(\left\{{}\begin{matrix}u:v=4:1\\50u+52v=6,3\end{matrix}\right.\)

=> u = 0,1 (mol); v = 0,025 (mol)

Kết tủa gồm \(\left\{{}\begin{matrix}CAg\equiv C-C\equiv CAg:0,1\left(mol\right)\\CAg\equiv C-CH=CH_2:0,025\left(mol\right)\end{matrix}\right.\)

=> m = 0,1.264 + 0,025.159 = 30,375 (g)

2Fe+3Cl₂-to>2FeCl₃

FeCl₃+3NaOH ->3NaCl+Fe(OH)₃

2Fe(OH)₃-to>Fe₂O₃+3H₂O

Fe₂O₃+3H₂-to>2Fe+3H₂O

3Fe+2O₂-to>Fe₃O₄