Ta có: \(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\)

a. PTHH: 2Na + 2H₂O ---> 2NaOH + H₂↑

b. Ta có: \(n_{H_2O}=\dfrac{97,8}{18}=5,43\left(mol\right)\)

Ta thấy: \(\dfrac{0,1}{2}< \dfrac{5,43}{2}\)

=> H₂O dư.

Theo PT: \(n_{H_2}=\dfrac{1}{2}.n_{Na}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)

=> \(V_{H_2}=0,05.22,4=1,12\left(lít\right)\)

c. Ta có: \(m_{dd_{NaOH}}=2,3+97,8=100,1\left(g\right)\)

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

=> \(m_{NaOH}=0,1.40=4\left(g\right)\)

=> \(C_{\%_{NaOH}}=\dfrac{4}{100,1}.100\%=3,996\%\)

\(n_{Na}=\dfrac{6,9}{23}=0,3\left(mol\right)\\ 2Na+2H_2O\rightarrow2NaOH+H_2\\ n_{H_2}=\dfrac{0,3}{2}=0,15\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\\ b,m_{ddsaup.ứ}=m_{Na}+m_{H_2O}-m_{H_2}=6,9+100-0,15.2=106,6\left(g\right)\)

\(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\) 
           0,4       0,8        0,4           0,4
\(a,V_{H_2}=0,4.22,4=8,96\left(l\right)\\ b,C\%_{HCl}=\dfrac{0,8.36,5}{150}.100\%=19,5\%\\ c,m_{\text{dd}}=26+150-\left(0,4.2\right)=175,2\left(g\right)\\ C\%_{ZnCl_2}=\dfrac{0,4.136}{175,2}.100\%=31\%\)

\(a)2Na+2H_2O\rightarrow2NaOH+H_2\\ b)n_{Na}=\dfrac{6,9}{23}=0,3mol\\2Na+2H_2O\rightarrow2NaOH+H_2\)

0,3            0,3          0,3            0,15

\(V_{H_2}=0,15.24,79=3,7455l\\ c)m_{NaOH}=0,3.40=12g\\ d)C_{\%NaOH}=\dfrac{12}{6,9+100-0,15.2}\cdot100=11,19\%\)

Sửa đề 2,6 -> 2,3 gam

a) $2Na + 2H_2O \to 2NaOH + H_2$

n Na = 2,3/23 = 0,1(mol)

n H2 = 1/2 n Na = 0,05(mol)

=> V H2 = 0,05.22,4 = 1,12(lít)

b) n NaOH = n Na = 2,3/23 = 0,1(mol)

=> CM NaOH = 0,1/0,2 = 0,5M

`n_[Zn]=13/65=0,2(mol)`

`Zn + 2HCl -> ZnCl_2 + H_2 \uparrow`

`0,2`  `0,4`                            `0,2`           `(mol)`

`a)V_[H_2]=0,2.22,4=4,48(l)`

`b)C%_[HCl]=[0,4.36,5]/150 .100~~9,73%`

`a)PTHH:`

  `Zn + 2HCl -> ZnCl_2 + H_2`

`0,02`                       `0,02`    `0,02`        `(mol)`

`n_[Zn]=[1,3]/65=0,02(mol)`

`b)V_[H_2]=0,02.22,4=0,448(l)`

`c)C%_[ZnCl_2]=[0,02.136]/[1,3+50-0,02.2].100~~5,31%`

\(n_{Zn}=\dfrac{1,3}{65}=0,02\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

0,02   0,04         0,02       0,02  ( mol )

\(V_{H_2}=0,02.22,4=0,448\left(l\right)\)

\(C\%_{ZnCl_2}=\dfrac{0,02.136}{1,3+50-0,02.2}.100=5,3\%\)

\(n_{Zn}=\dfrac{1,3}{65}=0,02\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\) 
         0,02                 0,02           0,02 
\(V_{H_2}=0,02.22,4=0,448\left(l\right)\\ m_{\text{dd}}=1,3+50-\left(0,02.2\right)=51,26\left(g\right)\\ C\%_{ZnCl_2}=\dfrac{0,02.136}{51,26}.100\%=5,3\%\)