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

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

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

a)

\(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)

PTHH: 2Na + 2H₂O --> 2NaOH + H₂

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

=> Chất tan trong dd X là NaOH

m_NaOH = 0,2.40 = 8 (g)

m_{dd sau pư} = 4,6 + 59,6 - 0,1.2 = 64 (g)

=> \(C\%=\dfrac{8}{64}.100\%=12,5\%\)

b) 

PTHH: CuO + H₂ --t^o--> Cu + H₂O

                       0,1------>0,1

=> m_Cu = 0,1.64 = 6,4 (g)

a)

\(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)

PTHH: 2Na + 2H₂O --> 2NaOH + H₂

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

=> Chất tan trong dd X là NaOH

m_NaOH = 0,2.40 = 8 (g)

m_{dd sau pư} = 4,6 + 59,6 - 0,1.2 = 64 (g)

=> \(C\%=\dfrac{8}{64}.100\%=12,5\%\)

b) 

PTHH: CuO + H₂ --t^o--> Cu + H₂O

                       0,1------>0,1

=> m_Cu = 0,1.64 = 6,4 (g)

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ a.Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{H_2}=n_{MgCl_2}=n_{Mg}=0,2\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ b.n_{HCl}=0,2.2=0,4\left(mol\right)\\ m_{ddHCl}=\dfrac{0,4.36,5.100}{7,3}=200\left(g\right)\\ c.m_{ddsau}=4,8+200-0,2.2=204,4\left(g\right)\\ C\%_{ddMgCl_2}=\dfrac{0,2.95}{204,4}.100\approx9,295\%\\ d.V_{ddHCl}=\dfrac{200}{1,05}=\dfrac{4000}{21}\left(ml\right)=\dfrac{4}{21}\left(l\right)\\ C_{MddHCl}=\dfrac{0,4}{\dfrac{4}{21}}=2,1\left(M\right)\)

Ta có: \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PTHH: 2Al + 6HCl ---> 2AlCl₃ + 3H₂.

Theo PT: \(n_{Al}=\dfrac{2}{3}.n_{H_2}=\dfrac{2}{3}.0,25=\dfrac{1}{6}\left(mol\right)\)

=> \(a=m_{Al}=\dfrac{1}{6}.27=4,5\left(g\right)\)

Theo PT: n_HCl = \(3.n_{Al}=3.\dfrac{1}{6}=0,5\)(mol)

=> m_HCl = 0,5 . 36,5 = 18,25(g)

=> \(C\%_{HCl}=\dfrac{m_{ct_{HCl}}}{m_{dd_{HCl}}}.100\%=\dfrac{18,25}{400}.100\%=4,5625\%\) 

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

PTHH: 2Al + 6HCl → 2AlCl₃ + 3H₂

Mol:    0,5/3     0,5                       0,25

\(m_{Al}=\dfrac{0,5}{3}.27=4,5\left(g\right)\)

\(C\%_{ddHCl}=\dfrac{0,5.36,5.100\%}{400}=4,5625\%\)

Tiếp: 

Ta có: \(m_{dd_{AlCl_3}}=4,5+400=404,5\left(g\right)\)

Theo PT: \(n_{AlCl_3}=n_{Al}=\dfrac{1}{6}\left(mol\right)\)

=> \(m_{AlCl_3}=\dfrac{1}{6}.133,5=22,25\left(g\right)\)

=> \(C\%_{AlCl_3}=\dfrac{22,25}{404,5}.100\%\approx5,5\%\)

Ta có: \(n_{Na}=\dfrac{9,2}{23}=0,4\left(mol\right)\)

\(PTHH:2Na+2H_2O--->2NaOH+H_2\uparrow\)

a. Theo PT: \(n_{H_2}=\dfrac{1}{2}.n_{Na}=\dfrac{1}{2}.0,4=0,2\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(lít\right)\)

b. Theo PT: \(n_{NaOH}=n_{Na}=0,4\left(mol\right)\)

\(\Rightarrow m_{NaOH}=0,4.40=16\left(g\right)\)

Ta có: \(m_{dd_{NaOH}}=9,2+200-0,2.2=208,8\left(g\right)\)

\(\Rightarrow C_{\%_{NaOH}}=\dfrac{16}{208,8}.100\%=7,66\%\)

1)

a) Fe + 2HCl --> FeCl₂ + H₂

b) \(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

           0,15->0,3--->0,15-->0,15

=> V_H2 = 0,15.22,4 = 3,36 (l)

c) m_{dd sau pư} = 8,4 + 250 - 0,15.2 = 258,1 (g)

=> \(C\%_{FeCl_2}=\dfrac{0,15.127}{258,1}.100\%=7,38\%\)

2)

a) Zn + 2HCl --> ZnCl₂ + H₂

b) \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

            0,2-->0,4---->0,2--->0,2

=> V_H2 = 0,2.22,4 = 4,48 (l)

m_ZnCl2 = 0,2.136 = 27,2 (g)

c) \(C_{M\left(dd.HCl\right)}=\dfrac{0,4}{0,2}=2M\)

d) 

PTHH: A + 2HCl --> ACl₂ + H₂

           0,2<--0,4

=> \(M_A=\dfrac{4,8}{0,2}=24\left(g/mol\right)\)

=> A là Mg(Magie)

a) 

\(n_{Zn}=\dfrac{16,25}{65}=0,25\left(mol\right)\)

PTHH: Zn + H₂SO₄ --> ZnSO₄ + H₂

         0,25-->0,25------------->0,25

=> V_H2 = 0,25.22,4 = 5,6 (l)

b) \(C_{M\left(dd.H_2SO_4\right)}=\dfrac{0,25}{0,3}=\dfrac{5}{6}M\)

c) \(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂ --t^o--> 2Fe + 3H₂O

Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,25}{3}\) => Fe₂O₃ dư, H₂ hết

PTHH: Fe₂O₃ + 3H₂ --t^o--> 2Fe + 3H₂O

           \(\dfrac{0,25}{3}\) <--0,25----->\(\dfrac{0,5}{3}\)

=> \(m=32-\dfrac{0,25}{3}.160+\dfrac{0,5}{3}.56=28\left(g\right)\)

PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

đb:       0,25

a) số mol của Zn là: \(n_{Zn}=\dfrac{m_{Zn}}{M_{Zn}}=\dfrac{16,25}{65}=0,25\left(mol\right)\)

Theo PTHH, ta có: \(n_{H_2}=\dfrac{0,25\cdot1}{1}=0,25\left(mol\right)\)

Thể tích của H₂ ở đktc là: \(V_{H_2\left(đktc\right)}=n_{H_2}\cdot22,4=0,25\cdot22,4=5,6\left(l\right)\)

2 câu còn lại mk chịu

em không biết

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