a)\(PTHH:2Na+2H_2O\xrightarrow[]{}2NaOH+H_2\)

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

\(PTHH:2Na+2H_2O\xrightarrow[]{}2NaOH+H_2\)

tỉ lệ        :2          2             2               1(mol)

số mol   :0,2       0,2          0,2             0,1

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

c)\(m_{NaOH}=0,2.40=8\left(g\right)\)

$a\big)$

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

$CH_3COOH+Na\to CH_3COONa+\dfrac{1}{2}H_2$

Theo PT: $n_{CH_3COOH}=n_{Na}=0,2(mol)$

$\to m_{CH_3COOH}=0,2.60=12(g)$

$b\big)$

Theo PT: $n_{H_2}=\dfrac{1}{2}n_{Na}=0,1(mol)$

$\to V_{H_2(đktc)}=0,1.22,4=2,24(l)$

giúp tuiii

a )

n_Mg = \(\dfrac{7,2}{24}\) 0,3 ( mol )
Mg + H₂SO₄ -> MgSO₄ + H₂

Theo pt : m_mgSO4 = 0,3.120 = 36 ( g )

b ) 

Theo pt : n_H2 = n_Mg = 0,3 ( mol )

-> V_{H2( đktc )} = 0,3.22,4 = 6,72 ( l )

a)

$M + 2HCl \to MCl_2 + H_2$
$n_{HCl} = 0,3.1 = 0,3(mol)$

Theo PTHH : $n_M = \dfrac{1}{2}n_{HCl} = 0,15(mol)$

$\Rightarrow M = \dfrac{3,6}{0,15} = 24(Mg)$

b)

$n_{MgCl_2} = n_{Mg} = 0,15(mol)$
$m_{MgCl_2} = 0,15.95 = 14,25(gam)$

c) $n_{H_2} = n_{Mg} = 0,15(mol)$
$V_{H_2} = 0,15.22,4 = 3,36(lít)$

a/ PTHH:2Al + 6HCl ===> 2AlCl3 + 3H2

n_Al = 5,4 / 27 = 0,2 mol

=> n_H2 = 0,3 mol

=> m_H2 = 0,3 x 2 = 0,6 gam

=> V_H2(đktc) = 0,3 x 22,4 = 6,72 lít

b/ => n_AlCl3 = 0,3 mol

=> m_AlCl3 = 0,2 x 133,5 = 26,7 gam

muối nhôm clorua là AlCl₃

a)

\(n_{Al}=0,2\left(mol\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(V_{H_2}=6,72\) ( lít )

\(m_{H_2}=0,6\left(g\right)\)

a) Zn + 2HCl → ZnCl₂ + H₂

b) nZn = 2 mol = nZnCl₂ = nH₂

=> mZnCl₂ = 2 .136 = 272 gam

c) nH₂ = 2.22,4 = 44,8 lít

n_Zn = 130 / 65 = 2 (mol) 

Zn + 2HCl => ZnCl2 + H2 

2____________2_____2 

m_ZnCl2 = 2 * 136 = 272 (g) 

V_H2 = 2 * 22.4 = 44.8 (l) 

dit me may fuck sex

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

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

0,2     0,2            0,2           0,2

a)\(V_{H_2}=0,2\cdot22,4=4,48l\)

b)\(m_{ZnSO_4}=0,2\cdot161=32,2g\)

  \(m_{ddZnSO_4}=30+200-0,2\cdot2=229,6g\)

  \(C\%=\dfrac{m_{ct}}{m_{dd}}\cdot100\%=\dfrac{32,2}{229,6}\cdot100\%=14,02\%\)

c)\(n_{CuO}=\dfrac{24}{80}=0,3mol\)

   \(CuO+H_2\rightarrow Cu+H_2O\)

   0,3        0,2     0,2

   \(m_{rắn}=m_{Cu}=0,2\cdot64=12,8g\)

C%=mctmdd⋅100%=19,6200⋅100%=9,8%C%=mctmdd⋅100%=19,6200⋅100%=9,8%

c)

bn mik hỏi tí aj
đoạn mdd ZnSO4 ấy, 30 ở đâu ra v ạ

\(a,n_{Na}=\dfrac{m_{Na}}{M_{Na}}=\dfrac{4,6}{23}=0,2\left(mol\right)\\ n_{Cl_2}=\dfrac{V_{Cl_2\left(đktc\right)}}{22,4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ PTHH:2Na+Cl_2\rightarrow2NaCl\\ Vì:\dfrac{0,2}{2}< \dfrac{0,2}{1}\Rightarrow Cl_2dư\\ \Rightarrow n_{Cl_2\left(dư\right)}=0,2-\dfrac{0,2}{2}=0,1\left(mol\right)\\ \Rightarrow m_{Cl_2\left(dư\right)}=0,1.71=7,1\left(g\right)\\ b,n_{NaCl}=n_{Na}=0,2\left(mol\right)\\ \Rightarrow m_{NaCl}=58,5.0,2=11,7\left(g\right)\)

a) $2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$

b) $n_{Al} = 0,45(mol) ; n_{H_2SO_4} =\dfrac{219}{980} (mol)$

Ta thấy : 

$n_{Al} : 2 > n_{H_2SO_4} : 3$ nên Al dư

Theo PTHH : 

$n_{Al\ pư} = \dfrac{2}{3}n_{H_2SO_4} = \dfrac{73}{490} (mol)$
$m_{Al\ dư} = 12,15 - \dfrac{73}{490}.27 = 8,127(gam)$

c) $n_{Al_2(SO_4)_3} = \dfrac{1}{3}n_{H_2SO_4} = \dfrac{73}{930}(mol)$
$m_{muối} = \dfrac{73}{930}.342 = 25,48(gam)$

d) $V_{H_2} = \dfrac{219}{980}.22,4 = 5(lít)$

Hình như đề sai

a,\(n_{Al}=\dfrac{12,15}{27}=0,45\left(mol\right)\)

  \(m_{H_2SO_4}=109,5.20\%=21,9\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{21,9}{98}=0,2235\left(mol\right)\)

\(a.2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{Al}=\dfrac{12,15}{27}=0,45\left(mol\right)\\ n_{HCl}=\dfrac{20\%.109,5}{36,5}=0,6\left(mol\right)\\ b.Vì:\dfrac{0,45}{2}>\dfrac{0,6}{6}\\ \Rightarrow Aldư\\ n_{Al\left(dư\right)}=0,45-\dfrac{2}{6}.0,6=0,25\left(mol\right)\\ m_{Al\left(dư\right)}=0,25.27=6,75\left(g\right)\\ c.n_{AlCl_3}=\dfrac{2}{6}.0,6=0,2\left(mol\right)\\ m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\\ d.n_{H_2}=\dfrac{3}{6}.0,6=0,3\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\)