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Bài 1:

\(2Na+2H_2O\rightarrow2NaOH+H_2\\ n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ n_{Na}=n_{Na_2O}=0,2.2=0,4\left(mol\right)\\ a.m_{Na}=0,4.23=9,2\left(g\right)\\ b.C_{MddA}=\dfrac{0,4}{0,5}=0,8\left(M\right)\\ C\%_{ddA}=\dfrac{0,4.40}{500.1,2}.100\approx2,667\%\)

PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

            \(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

a) Ta có: \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)=n_{Zn}\)

\(\Rightarrow\%m_{Zn}=\dfrac{0,4\cdot65}{36,2}\cdot100\%\approx71,23\%\) \(\Rightarrow\%m_{Al_2O_3}=28,77\%\)

c) Ta có: \(n_{Al_2O_3}=\dfrac{36,2-0,4\cdot65}{102}=0,1\left(mol\right)\)

Theo PTHH: \(n_{HCl}=2n_{Zn}+6n_{Al_2O_3}=1,4\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{1,4\cdot36,5}{10\%}=511\left(g\right)\) \(\Rightarrow V_{ddHCl}=\dfrac{511}{1,1}\approx464,5\left(ml\right)=0,4645\left(l\right)\)

c) Theo PTHH: \(\left\{{}\begin{matrix}n_{ZnCl_2}=0,4\left(mol\right)\\n_{AlCl_3}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{ZnCl_2}}=\dfrac{0,4}{0,4645}\approx0,86\left(M\right)\\C_{M_{AlCl_3}}=\dfrac{0,2}{0,4645}\approx0,43\left(M\right)\end{matrix}\right.\)

 

 

24 tháng 3 2022

$a\big)$

$Zn+2CH_3COOH\to (CH_3COO)_2Zn+H_2$

$ZnO+2CH_3COOH\to (CH_2COO)_2Zn+H_2O$

Theo PT: $n_{Zn}=n_{H_2}=\frac{4,48}{22,4}=0,2(mol)$

$\to \%m_{Zn}=\frac{0,2.65}{21,1}.100\%\approx 61,61\%$

$\to \%m_{ZnO}=100-61,61=38,39\%$

$b\big)$

$n_{ZnO}=\frac{21,1-0,2.65}{81}=0,1(mol)$

Theo PT: $\sum n_{CH_3COOH}=2n_{Zn}+2n_{ZnO}=0,6(mol)$

$\to C_{M_{CH_3COOH}}=\dfrac{0,6}{\frac{200}{1000}}=3M$

24 tháng 3 2022

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2mol\)

\(Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\)

 0,2                                                                0,2     ( mol )

\(m_{Zn}=0,2.65=13g\)

\(\%m_{Zn}=\dfrac{13}{21,1}.100=61,61\%\)

\(\%m_{ZnO}=100\%-61,61\%=38,39\%\)

\(Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\)

0,2               0,4                                                  ( mol )

\(n_{ZnO}=\dfrac{21,1-13}{81}=0,1mol\)

\(ZnO+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2O\)

 0,1              0,2                                                        ( mol )

\(C_{M\left(CH_3COOH\right)}=\dfrac{0,4+0,2}{0,2}=3M\)

 

 

 

 

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

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

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

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

8 tháng 5 2022

nZn=1365=0,2molnZn=1365=0,2mol

Zn+H2SO4→ZnSO4+H2Zn+H2SO4→ZnSO4+H2

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

a)VH2=0,2⋅22,4=4,48lVH2=0,2⋅22,4=4,48l

b)mH2SO4=0,2⋅98=19,6gmH2SO4=0,2⋅98=19,6g

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

c)nCuO=2480=0,3molnCuO=2480=0,3mol

   CuO+H2→Cu+H2OCuO+H2→Cu+H2O

   0,3        0,2     0,2

   mrắn=mCu=0,2⋅64=12,8g.

26 tháng 12 2022

a)

Gọi $n_{Fe} = a(mol) ; n_{Al} =b (mol) \Rightarrow 56a + 27b = 11(1)$
$Fe + 2HCl \to FeCl_2 + H_2$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
Theo PTHH : $n_{H_2} = a + 1,5b = \dfrac{8,96}{22,4} = 0,4(2)$

Từ (1)(2) suy ra : a = 0,1 ; b = 0,2

$\%m_{Fe} = \dfrac{0,1.56}{11}.100\% = 50,9\%$
$\%m_{Al} = 100\% - 50,9\% = 49,1\%$

b) $n_{HCl} = 2n_{H_2} = 0,8(mol)$
$\Rightarrow C_{M_{HCl}} = \dfrac{0,8}{0,4} = 2M$

c)

$C_{M_{FeCl_2}} = \dfrac{0,1}{0,4} = 0,25M$
$C_{M_{AlCl_3}} =\dfrac{0,2}{0,4} = 0,5M$

9 tháng 3 2022

mình đg cần gấp

9 tháng 3 2022

a) nNa = 4,6/23 = 0,2 (mol)

PTHH: 2Na + 2H2O -> 2NaOH + H2

Mol: 0,2 ---> 0,2 ---> 0,2 ---> 0,1

VH2 = 0,1 . 22,4 = 2,24 (l)

b) CMNaOH = 0,2/0,1 = 2M

c) mH2O = 100 . 1 = 100 (g)

mNaOH = 0,2 . 40 = 8 (g)

mH2 = 0,1 . 2 = 0,2 (g)

mdd = 100 + 8 - 0,2 = 107,8 (g)

C%NaOH = 8/107,8 = 7,42%

25 tháng 12 2023

a, Ta có: 27nAl + 56nFe = 22 (1)

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

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=\dfrac{19,832}{24,79}=0,8\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,4\left(mol\right)\\n_{Fe}=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,4.27}{22}.100\%\approx49,09\%\\\%m_{Fe}\approx50,91\%\end{matrix}\right.\)

b, \(n_{HCl}=2n_{H_2}=1,6\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{1,6}{0,5}=3,2\left(M\right)\)

19 tháng 1 2022

$a)PTHH:2Al+6HCl\to 2AlCl_3+3H_2$

$n_{H_2}=\dfrac{5,04}{22,4}=0,225(mol)$

$\Rightarrow n_{Al}=0,15(mol)$

$\Rightarrow \%m_{Al}=\dfrac{0,15.27}{9,45}.100\%\approx 42,86\%$

$\Rightarrow \%m_{Cu}=100-42,86=57,14\%$

$b)$ Theo PT: $n_{HCl}=2n_{H_2}=0,45(mol)$

$\Rightarrow C_{M_{HCl}}=\dfrac{0,45.110\%}{0,5}=0,99M$