\(\text{2K + 2H2O -> 2KOH + H2}\)

\(\text{a) Ta có: nK=19,5/39=0,5 mol}\)

Theo ptpu: nH2=1/2nK=0,25 mol -> V H2=0,25.22,4=5,6 lít

b)

Dung dịch sau phản ứng chứa KOH -> nKOH=nK=0,5 mol -> mKOH=0,5.56=28 gam

BTKL: m dung dịch sau phản ứng=mK + mH2O - mH2

\(\text{=19,5+200-0,25.2=219 gam -> %KOH=28/219=12,7%}\)

a) m_HCl = 14,6% . 200 = 29,2 ( g )

⇒ n_HCl = \(\dfrac{m}{M}\) = \(\dfrac{29,2}{36,5}\) = 0,8 ( mol )

PTHH : Zn + 2HCl → ZnCl₂ + H₂

             0,4    0,8          0,4      0,2    ( mol )

Theo pt : nH₂ = 0,4 mol

          ⇒ VH_2(đktc) = nH₂ . 22.4 = 0,4 . 22,4 = 8,96 ( l )

b) Theo pt : m_Zn = n.M = 0,4 . 65 = 26 ( g )

c) mH₂ = n.M = 0,4 . 2 = 0,8 ( g )

Theo pt : mZnCl₂ = n.M = 0,4 . 136 = 54,4 ( g )

         ⇒ m_dd(sau) = 200 + 26 - 0,8 = 225,2 ( g )

         ⇒ C%ZnCl₂ = \(\dfrac{54,4}{225,2}\) . 100% ≃ 24,16%

\(n_{Al} = \dfrac{10,8}{27} = 0,4(mol)\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ n_{AlCl_3} = n_{Al} = 0,4(mol)\\ m_{AlCl_3} = 0,4.133,5 = 53,4(gam)\\ n_{HCl} =3 n_{Al} = 1,2(mol)\\ C\%_{HCl}= \dfrac{1,2.36,5}{200}.100\% = 21,9\%\)

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

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

\(4.a/n_{Al}=\dfrac{5,4}{27}=0,2mol\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,2       0,3                  0,1                  0,3

\(V_{H_2}=0,3.24,79=7,437l\\ b/C_{\%H_2SO_4}=\dfrac{0,3.98}{150}\cdot100=19,6\%\\ c/m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2g\)

\(5.a/n_{MgO}=\dfrac{4}{40}=0,1mol\\ MgO+2HCl\rightarrow MgCl_2+H_2O\)

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

\(C_{\%HCl}=\dfrac{0,2.36,5}{200}\cdot100=3,65\%\\ b/C_{\%MgCl_2}=\dfrac{0,1.95}{200+4}\cdot100=4,66\%\\ c/NaOH+HCl\rightarrow NaCl+H_2O\\ n_{NaOH}=n_{HCl}=0,2mol\\ V_{NaOH}=\dfrac{0,2}{1}=0,2l=200ml\)

a) \(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

            0,4-->0,8----->0,4--->0,4

=> V_H2 = 0,4.22,4 = 8,96 (l)

b) m_ZnCl2 = 0,4.136 = 54,4 (g)

c) \(C\%=\dfrac{0,8.36,5}{200}.100\%=14,6\%\)

nK = 7,8 : 39 = 0,2 (mol) 
pthh : 2K + 2H2O ---> 2KOH + H2
          0,2--------------------------> 0,1 (mol) 
=> VO2 = 0,1 . 22,4 = 2,24 (l)

a) \(PTHH:2K+2H_2O\rightarrow2KOH+H_2\)

b) \(n_K=\dfrac{m}{M}=\dfrac{7,8}{39}=0,2\left(mol\right)\)

Theo pthh: \(n_{H_2}=\dfrac{1}{2}n_K=\dfrac{1}{2}.0,2=0,1\left(mol\right)\)

\(V_{H_2}=n.22,4=0,1.22,4=2,24\left(l\right)\)

\(4.\\ a/n_{Al}=\dfrac{5,4}{27}=0,2mol\\2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

\(0,2......0,3............0,1...............0,3\)

\(V_{H_2}=0,3.24,79=7,437l\\ b.C_{\%H_2SO_4}=\dfrac{0,3.98}{150}\cdot100=19,6\%\\ c.m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2g\)

\(7.\\a) n_{H_2SO_4}=\dfrac{100.9,8}{100.98}=0,1mol\\ ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\\ n_{ZnO}=n_{ZnSO_4}=n_{H_2SO_4}=0,1mol\\ m_{ZnO}=0,1.81=8,1g\\ b)C_{\%ZnSO_4}=\dfrac{0,1.161}{8,1+100}\cdot100=14,8\%\)

\(9.\\ a)n_{H_2}=\dfrac{2,479}{24,79}=0,1mol\\ Mg+2HCl\rightarrow MgCl_2+H_2\)

\(0,1.....0,2.........0,1........0,1\)

\(m_{Mg}=0,1.24=2,4g\\ m_{MgO}=8,4-2,4=6g\\ b)n_{MgO}=\dfrac{6}{40}=0,15mol\\ MgO+2HCl\rightarrow MgCl_2+H_2O\\ n_{HCl}=0,15.2+0,2=0,5mol\\ m_{ddHCl}=\dfrac{0,5.36,5}{3,65}\cdot100=500g\)