a) \(3NaOH+FeCl_3\xrightarrow[]{}3NaCl+Fe\left(OH\right)_3\)

(mol)  0,9...............0,3.............................0,3

\(2Fe\left(OH\right)_3\xrightarrow[]{t^o}Fe_2O_3+3H_2O\)

(mol) 0,3...............0,15

200ml=0,2l

\(n_{FeCl_3}=0,2.1,5=0,3\left(mol\right)\)

\(V_{NaOH}=\dfrac{0,9}{2}=0,45\left(l\right)\)

b) \(m_{Fe\left(OH\right)_3}=0,3.107=32,1\left(g\right)\)

\(m_{Fe_2O_3}=0,15.160=24\left(g\right)\)

a) \(3Cl_2+2Fe\xrightarrow[]{t^o}2FeCl_3\\ Cl_2+2NaOH\xrightarrow[]{}NaCl+NaClO+H_2O\)

b) 

TN1: không xảy ra phản ứng

TN2: \(Fe+CuSO_4\xrightarrow[]{}FeSO_4+Cu\)

*Nhận xét: sắt hoạt động hóa học mạnh hơn đồng

a)           \(2Al+6HCl\xrightarrow[]{}2AlCl_3+3H_2\)

(mol)     0,05..........0,15........................0,075

- Hiện tượng: bột aluminium tan dần, có khí không màu thoát ra

b) 150ml=0,15l

\(n_{HCl}=0,15.1=0,15\left(mol\right)\)

\(m_{Al}=0,05.27=1,35\left(g\right)\)

\(V_{H_2}=0,075.22,4=1,68\left(l\right)\)

a) PT: \(NaCl+AgNO_3\xrightarrow[]{}NaNO_3+AgCl\)

(mol)  0,04..........0,04...........0,04.............0,04

\(n_{NaCl}=\dfrac{2,34}{58,5}=0,04\left(mol\right)\)

\(m_{AgCl}=143,5.0,04=5,74\left(g\right)\)

b) \(m_{AgNO_3}=170.0,04=6,8\left(g\right)\)

c) Theo BTKL: \(m_{NaCl}+m_{ddAgNO_3}=m_{ddsau}+m_{AgCl}\)

⇒ \(m_{ddsau}=2,34+\dfrac{6,8.100}{3,4}-5,74=196,6\left(g\right)\)

\(C\%_{NaNO_3}=\dfrac{85.0,04}{196,6}.100=1,73\%\)

Gọi CT của ankan là \(C_nH_{2n+2}\), của anken là \(C_mH_{2m}\) (\(1\le n\le4\);  \(2\le m\le4\))

TN1: \(C_mH_{2m}+H_2\xrightarrow[Ni]{t^o}C_mH_{2m+2}\)

(mol)  0,005........0,005..........0,005

TN2: \(C_mH_{2m}+Br_2\xrightarrow[]{}C_mH_{2m}Br_2\)

Khi cho lượng khí lội qua brom dư thì thể tích giảm

⇒ Anken từ TN1 còn dư và \(H_2\) phản ứng hết

 \(\Rightarrow V_{H_2}=560-448=112\left(ml\right)=0,112\left(l\right)\)

và \(V_{C_mH_{2m}dư}=448-280=168\left(ml\right)=0,168\left(l\right)\)

\(n_{H_2}=\dfrac{0,112}{22,4}=0,005\left(mol\right)\)

\(n_{C_mH_{2m}\left(2\right)}=\dfrac{0,168}{22,4}=0,0075\left(mol\right)\)

\(n_{C_mH_{2m}banđầu}=0,005+0,0075=0,0125\left(mol\right)\)

\(n_{C_nH_{2n+2}}=\dfrac{0,56}{22,4}-0,0125-0,005=0,0075\left(mol\right)\)

hh sau gồm \(C_nH_{2n+2}\) và \(C_mH_{2m+2}\)

 \(M_{hh}=d.M_{H_2}\)

⇒ \(\dfrac{0,0075\left(14n+2\right)+0,005\left(14m+2\right)}{0,0075+0,005}=17,8.2\)

⇒ \(0,105n+0,07m=0,42\)

⇒ \(3n+2m=12\)

\(\begin{matrix}n&1&2&3&4&\\m&4,5&3&1,5&0&\\KQ&X&chọn&X&X&\end{matrix}\)

⇒ CT của ankan là \(C_2H_6\), của anken là \(C_3H_6\)

Trong cùng đk nhiệt độ, áp suất thì \(\%^V\) cũng là \(\%^n\)

 \(\%^V_{C_2H_6}=\dfrac{0,0075}{0,025}.100=30\%\)

\(\%^V_{C_3H_8}=\dfrac{0,0125}{0,025}.100=50\%\)

a) \(C_2H_4+Br_2\xrightarrow[]{}C_2H_4Br_2\)

b) \(n_{C_2H_4Br_2}=n_{C_2H_4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

\(m_{C_2H_4Br_2}=0,5.188=94\left(g\right)\)

c) \(n_{Br_2}=n_{C_2H_4}=0,5\left(mol\right)\)

500ml=0,5l

\(C_{M_{Br_2}}=\dfrac{0,5}{0,5}=1\left(M\right)\)

          \(BaCl_2+H_2SO_4\xrightarrow[]{}BaSO_4+2HCl\)

(mol)    0,05.............0,05..............0,05..........0,1

\(n_{BaCl_2}=\dfrac{104.10}{100.208}=0,05\left(mol\right)\)

\(m_{H_2SO_4}=0,05.98=4,9\left(g\right)\)

Theo BTKL: \(m_{ddH_2SO_4}+m_{ddBaCl_2}=m_{ddsau}+m_{BaSO_4}\)

⇒ \(m_{ddsau}=\dfrac{4,9.100}{7}+104-0,05.233=162,35\left(g\right)\)

\(C\%_{HCl}=\dfrac{0,1.36,5}{162,35}.100=2,25\%\)

20. \(2K+H_2SO_4\xrightarrow[]{}K_2SO_4+H_2\)

21. \(2Na+H_2SO_4\xrightarrow[]{}Na_2SO_4+H_2\)

23. \(3KOH+AlCl_3\xrightarrow[]{}Al\left(OH\right)_3+3KCl\)

24. \(6NaOH+Al_2\left(SO_4\right)_3\xrightarrow[]{}3Na_2SO_4+2Al\left(OH\right)_3\)

25. \(H_2SO_4+Ba\left(OH\right)_2\xrightarrow[]{}BaSO_4+2H_2O\)

26. \(3HCl+Fe\left(OH\right)_3\xrightarrow[]{}FeCl_3+3H_2O\)

27. \(2HCl+Zn\left(OH\right)_2\xrightarrow[]{}ZnCl_2+2H_2O\)

28. \(2HCl+Mg\left(OH\right)_2\xrightarrow[]{}MgCl_2+2H_2O\)

29. \(2HCl+Fe\left(OH\right)_2\xrightarrow[]{}FeCl_2+2H_2O\)

30. \(H_2SO_4+Fe\left(OH\right)_2\xrightarrow[]{}FeSO_4+2H_2O\)

31. \(3H_2SO_4+2Fe\left(OH\right)_3\xrightarrow[]{}Fe_2\left(SO_4\right)_3+6H_2O\)

32. \(H_2SO_4+Mg\left(OH\right)_2\xrightarrow[]{}MgSO_4+2H_2O\)

33. \(2Al+6HCl\xrightarrow[]{}2AlCl_3+3H_2\)

34. \(2Al+3H_2SO_4\xrightarrow[]{}Al_2\left(SO_4\right)_3+3H_2\)

a) \(Mg+H_2SO_4\xrightarrow[]{}MgSO_4+H_2\)

(mol) 0,2.......0,2............0,2...............0,2

b) \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

 \(V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)

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

d) \(2NaOH+MgSO_4\xrightarrow[]{}Na_2SO_4+Mg\left(OH\right)_2\)

(mol)  0,4.............0,2..................................0,2

\(m_{ddNaOH}=\dfrac{0,4.40.100}{5}=320\left(g\right)\)

\(m_{Mg\left(OH\right)_2}=0,2.58=11,6\left(g\right)\)

a) \(Mg+H_2SO_4\xrightarrow[]{}MgSO_4+H_2\\ MgO+H_2SO_4\xrightarrow[]{}MgSO_4+H_2O\)

b) \(n_{H_2SO_4}=\dfrac{200.24,5}{100.98}=0,5\left(mol\right)\)

Gọi x, y là số mol Mg, MgO

⇒ \(n_{H_2SO_4}=x+y=0,5\left(mol\right)\)

Ta có hpt: \(\left\{{}\begin{matrix}24x+40y=16,8\\x+y=0,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,3\end{matrix}\right.\)

\(n_{H_2}=n_{Mg}=0,2\left(mol\right)\)

Theo BTKL: \(m_{hh}+m_{ddH_2SO_4}=m_{ddsau}+m_{H_2}\)

⇒ \(m_{ddsau}=16,8+200-0,2.2=216,4\left(g\right)\)

\(m_{MgSO_4}=120.\left(0,2+0,3\right)=60\left(g\right)\)

\(C\%_{MgSO_4}=\dfrac{60}{216,4}.100=27,73\%\)