\(2Na+2H_2O\rightarrow2NaOH+H_2\)

\(2K+2H_2O\rightarrow2KOH+H_2\)

Gọi: \(\left\{{}\begin{matrix}n_{Na}=x\left(mol\right)\\n_K=y\left(mol\right)\end{matrix}\right.\) ⇒ 23x + 39y = 12,4 (1)

Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{Na}+\dfrac{1}{2}n_K=\dfrac{1}{2}x+\dfrac{1}{2}y=\dfrac{2,24}{22,4}=0,1\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=-0,2875\\y=0,4875\end{matrix}\right.\)

Đến đây thì ra số mol âm, bạn xem lại đề nhé.

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PT: \(2K+2H_2O\rightarrow2KOH+H_2\)

\(K_2O+H_2O\rightarrow2KOH\)

Theo PT: \(n_K=2n_{H_2}=0,2\left(mol\right)\Rightarrow m_K=0,2.39=7,8\left(g\right)\)

\(\Rightarrow m_{K_2O}=12,4-7,8=4,6\left(g\right)\)

\(2Na+2H_2O\rightarrow2NaOH+H_2\\ Na_2O+H_2O\rightarrow2NaOH\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{Na}=2.0,3=0,6\left(mol\right)\\ a,m_{Na}=0,6.23=13,8\left(g\right)\\ m_{Na_2O}=26,2-13,8=12,4\left(g\right)\\b, n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\\ n_{NaOH\left(tổng\right)}=n_{Na}+2.n_{Na_2O}=0,6+\dfrac{12,4}{62}=0,8\left(mol\right)\\ m_{c.tan}=m_{NaOH}=0,8.40=32\left(g\right)\\ c,m_{ddNaOH}=m_{hh}+m_{H_2O}-m_{H_2}=26,2+200-0,3.2=225,6\left(g\right)\\ C\%_{ddNaOH}=\dfrac{32}{225,6}.100\approx14,185\%\)

Anh có làm rồi nha!

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Theo PT: \(n_{Fe}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)

\(\Rightarrow m_{Cu}=12-5,6=6,4\left(g\right)\)

Nhận thấy rừng `Cu` không tác dụng với `HCl` nên toàn bộ lượng `H_2` là do `Fe`

\(n_{H_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ PTHH:Fe+2HCl->FeCl_2+H_2\)

tỉ lệ         1    :       2      :         1     :      1

n(mol)    0,1<------------------------------0,1

\(m_{Fe}=n\cdot M=0,1\cdot56=5,6\left(g\right)\\ m_{Cu}=12-5,6=6,4\left(g\right)\)

2Na+2H2O->2NaOH+H2

0,5-----0,5-----------0,5----0,25

Na2O+H2O->2NaOH

0,1--------0,1-----------0,2

n H2=0,25 mol

=>m Na =0,5.23=11,5g

=>m Na2O=6,2g=>n Na2O=0,1 mol

=>m NaOH=0,7.40=28g

=>VH2O=0,6.22,4=13,44l

giúp mình với

\(n_{H_2}=\dfrac{5,6}{22,4}=0,25mol\)

\(2Na+2H_2O\rightarrow2NaOH+H_2\)

0,5                                        0,25

a)\(m_{Na}=\left(\right)\cdot23=11,5g\)

\(m_{Na_2O}=17,7-11,5=6,2g\)

b)\(n_{Na_2O}=\dfrac{6,2}{62}=0,1mol\)

\(Na_2O+H_2O\rightarrow2NaOH\)

0,1                           0,2

\(\Rightarrow\Sigma n_{bazo}=0,5+0,2=0,7mol\Rightarrow m_{NaOH}=0,7\cdot40=28g\)

c)\(m_{H_2O}=\left(0,5+0,1\right)\cdot18=10,8g\)

Gọi số mol Na, Zn là a, b

=> 23a + 65b = 14,3 

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

- Nếu Zn tan hết

PTHH: 2Na + 2H₂O --> 2NaOH + H₂

______a-------------------->a---->0,5a

2NaOH + Zn --> Na₂ZnO₂ + H₂

__2b<----b-------------------->b

=> \(\left\{{}\begin{matrix}2b\le a\\0,5a+b=14,3\end{matrix}\right.\) => Loại 

=> Zn không tan hết => NaOH hết

PTHH: 2Na + 2H₂O --> 2NaOH + H₂

______a------------------->a---->0,5a

2NaOH + Zn --> Na₂ZnO₂ + H₂

_a--------------------------->0,5a

=> 0,5a + 0,5a = 0,1

=> a = 0,1

=> m_Na = 0,1.23 = 2,3 (g)

=> m_Zn = 14,3 - 2,3 = 12(g)

a)

2C₂H₅OH + 2Na --> 2C₂H₅ONa + H₂

2C₆H₅OH + 2Na --> 2C₆H₅ONa + H₂

b) 

Gọi số mol C₂H₅OH, C₆H₅OH là a, b (mol)

=> 46a + 94b = 14 (1)

PTHH: 2C₂H₅OH + 2Na --> 2C₂H₅ONa + H₂

                    a---------------------------->0,5a

           2C₆H₅OH + 2Na --> 2C₆H₅ONa + H₂

                    b---------------------------->0,5b

=> 0,5a + 0,5b = \(\dfrac{2,24}{22,4}=0,1\) (2)

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

\(\left\{{}\begin{matrix}\%m_{C_2H_5OH}=\dfrac{0,1.46}{14}.100\%=32,857\%\\\%m_{C_6H_5OH}=\dfrac{0,1.94}{14}.100\%=67,143\%\end{matrix}\right.\)

c) 

PTHH: C₆H₅OH + 3HNO_3(đ) --H₂SO_4(đ)--> C₆H₂(NO₂)₃OH + 3H₂O

                  0,1---------------------------------->0,1

=> \(m_{C_6H_2\left(NO_2\right)_3OH}=0,1.229=22,9\left(g\right)\)