\(2K+2H_2O->2KOH+H_2\\ 2Na+2H_2O->2NaOH+H_2\\ Ca+2H_2O->Ca\left(OH\right)_2+H_2\\ Ba+2H_2O->Ba\left(OH\right)_2+H_2\\ 2Li+2H_2O->2LiOH+H_2\\ CaO+H_2O->Ca\left(OH\right)_2\\ Na_2O+H_2O->2NaOH\\ BaO+H_2O->Ba\left(OH\right)_2\\ P_2O_5+3H_2O->2H_3PO_4\\ SO_3+H_2O->H_2SO_4\)
\(O_2+2H_2-t^o->2H_2O \\ CuO+H_2-t^O->Cu+H_2O\\ Fe_2O_3+3H_2-t^O->2Fe+3H_2O\\ PbO+H_2-t^O->Pb+H_2O\\ Fe_3O_4+4H_2-t^O->3Fe+4H_2O\)

a/ 

\(2K+2H_2O\rightarrow2KOH+H_2\\ 2Na+2H_2O\rightarrow2NaOH+H_2\\ Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\\ Ba+2H_2O\rightarrow Ba\left(OH\right)_2+H_2\\ 2Li+H_2O\rightarrow2LiOH+H_2\\ CaO+H_2O\rightarrow Ca\left(OH\right)_2+H_2\\ Na_2O\rightarrow2NaOH\\ BaO+H_2O\rightarrow Ba\left(OH\right)_2\\ P_2O_5+3H_2O\rightarrow2H_3PO_4\\ SO_3+H_2O\rightarrow H_2SO_4\)

b/ 

\(2H_2+O_2\underrightarrow{t^o}2H_2O\\ CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ PbO+H_2\underrightarrow{t^o}Pb+H_2O\\ Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\\ Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)

\(n_{Ca}=\dfrac{8}{40}=0,2\left(mol\right)\)

PTHH: Ca + 2H₂O ---> Ca(OH)₂ + H₂

            0,2                                       0,2

           Fe₂O₃ + 3H₂ --t^o--> 2Fe + 3H₂O

                          0,2             \(\dfrac{2}{15}\)

\(\rightarrow\left\{{}\begin{matrix}V_{H_2}=0,2.22,4=4,48\left(l\right)\\m_{Fe}=\dfrac{2}{15}.56=\dfrac{112}{15}\left(g\right)\end{matrix}\right.\)

\(n_{Ca}=8:40=0,2\left(mol\right)\)
pthh : \(Ca+2H_2O->Ca\left(OH\right)_2+H_2\)
          0,2                                    0,2 
=> \(V_{H_2}=0,2.22,4=4,48\left(L\right)\)
\(pthh:Fe_2O_3+3H_2-t^o->2Fe+3H_2O\)
                           0,2                 \(\dfrac{2}{15}\)
=> \(m_{Fe}=\dfrac{2}{15}.56=\dfrac{112}{15}\left(G\right)\)

Gọi \(m_{CuSO_4\left(75^oC\right)}=a\left(g\right)\)

\(\rightarrow S_{CuSO_4\left(75^oC\right)}=\dfrac{a}{1654-a}.100=65,4\\ \Leftrightarrow a=654\left(g\right)\)

\(\rightarrow m_{H_2O}=1654-654=1000\left(g\right)\)

Giả sử mỗi ddbh có 100 g nước

\(\rightarrow m_{CuSO_4\left(\text{kết tinh}\right)}=65,4-10=55,4\left(g\right)\)

Mà thực tế có 1000 g nước 

\(\rightarrow m_{CuSO_4\left(\text{kết tinh}\right)}=\dfrac{1000}{100}.55,4=554\left(g\right)\)

\(n_{SO_3}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

PTHH: SO₃ + H₂O ---> H₂SO₄

           0,4                        0,4

=> \(m_{H_2SO_4}=0,4.98=39,2\left(g\right)\)

a) 2Na + 2H₂O --> 2NaOH + H₂

b) \(n_{H_2}=\dfrac{3,76}{22,4}=\dfrac{47}{280}\left(mol\right)\)

PTHH: 2Na + 2H₂O --> 2NaOH + H₂

           \(\dfrac{47}{140}\)<------------------\(\dfrac{47}{280}\)

=> \(m_{Na}=\dfrac{47}{140}.23=\dfrac{1081}{140}\left(g\right)\)

C

C

kiểm tra ko giúp dc nhé

hmmmmmm dạng bài tập hả 
cân bằng pthh 
tính theo pthh 
cách đọc tên phân loại axit bazo muối 

1) 4Al + 3O₂ --t^o--> 2Al₂O₃

2) 2NaOH + H₂SO₄ --> Na₂SO₄ + 2H₂O

3) 2Al + 6HCl --> 2AlCl₃ + 3H₂

4) Fe₂O₃ + 3H₂SO₄ --> Fe₂(SO₄)₃ + 3H₂O

5) P₂O₅ + 3H₂O --> 2H₃PO₄

6) 3Ca(OH)₂ + 2FeCl₃ --> 3CaCl₂ + 2Fe(OH)₃

19D

20B

21B

22A