a, Ta có: 27n_Al + 56n_Fe = 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, \(n_{HCl}=2n_{H_2}=1,6\left(mol\right)\)

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

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

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

Theo PT: \(n_{H_2}=n_{Zn}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)

c, \(n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\Rightarrow m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)

d, \(n_{HCl}=2n_{Zn}=0,4\left(mol\right)\Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{14,6}{7,3\%}=200\left(g\right)\)

⇒ m _{dd sau pư} = 13 + 200 - 0,2.2 = 212,6 (g)

\(\Rightarrow C\%_{ZnCl_2}=\dfrac{27,2}{212,6}.100\%\approx12,79\%\)

Gọi x,y,z lần lượt là số mol của Al,Mg,Zn

PT: 

2Al + 6HCl--->2AlCl3 + 3H2

x-------3x----------------------1,5x   mol

Mg + 2HCl--->MgCl2 + H2

y-----2y----------------------y    mol

Zn + 2HCl--->ZnCl2 + H2

z----2z--------------------z      mol

b.

Số mol H2: nH2=16,352/22,4=0,73 mol

1,5x+y+z=0,73

27x = 24y =>x=8y/9

=>7y/3 +z =0,73 (*)

27x + 24y + 65z=19,6

27x = 24y

=> 48y + 65z =19,6 (**)

Từ (*),(**)

=>y=0,27 => mMg =6,48 g

z=0,1=>mZn = 6,5 g

x=0,24=>mAl =6,48g

c.

nHCl =2nH2

=>nHCl =2.0,73=1,46 mol

=>V dd=1,46/2=0,73(l)

Gọi \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)

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

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

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

Theo pt: \(\Rightarrow\left\{{}\begin{matrix}3x+y=0,2\\27x+56y=5,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{19}{470}\\y=\dfrac{37}{470}\end{matrix}\right.\)

\(\%m_{Al}=\dfrac{\dfrac{19}{470}\cdot27}{5,5}\cdot100\%=19,84\%\)

\(\%m_{Fe}=100\%-19,84\%=80,16\%\)

a, Ta có: 27n_Al + 56n_Fe = 27,8 (1)

PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=\dfrac{17,353}{24,79}=0,7\left(mol\right)\left(2\right)\)

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

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2.27}{27,8}.100\%\approx19,42\%\\\%m_{Fe}\approx80,58\%\end{matrix}\right.\)

b, \(n_{H_2SO_4}=n_{H_2}=0,7\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,7}{0,5}=1,4\left(M\right)\)

a) Gọi $n_{Al} =a (mol) ; n_{Fe} = b(mol) \Rightarrow 27a + 56b = 13(1)$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
$Fe + 2HCl \to FeCl_2 + H_2$
Theo PTHH : $n_{H_2} = 1,5a + b = \dfrac{6,72}{22,4} = 0,3(2)$
Từ (1)(2) suy ra : $a = \dfrac{1}{15} ; b = 0,2$

$\%m_{Al} = \dfrac{ \dfrac{1}{15}.27}{13}.100\% = 13,8\%$

$\%m_{Fe} = 100\% - 13,8\% = 86,2\%$

b) $n_{HCl} = 2n_{H_2} = 0,3.2 = 0,6(mol)$
$\Rightarrow C_{M_{HCl}} = \dfrac{0,6}{0,15} = 4M$

c) $n_{muối} = m_{kim\ loại} + m_{HCl} - m_{H_2} = 13 + 0,6.36,5 - 0,3.2 = 34,3(gam)$

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

b) n Al = 8,1/27 = 0,3(mol)

Theo PTHH : 

n H2 = 3/2 n Al = 0,45(mol)

V H2 = 0,45.22,4 = 10,08(lít)

c) n AlCl3 = n Al = 0,3(mol)

m AlCl3 = 0,3.133,5 = 40,05(gam)

d) n HCl = 3n Al = 0,9(mol)

m dd HCl = 0,9.36,5/7,3% = 450(gam)

Sau phản ứng : 

m dd = 8,1 + 450  -0,45.2 = 457,2(gam)

C% AlCl3 = 40,05/457,2  .100% = 8,76%

a. PTHH:

\(BaO+2HCl--->BaCl_2+H_2O\left(1\right)\)

\(BaCO_3+2HCl--->BaCl_2+CO_2\uparrow+H_2O\left(2\right)\)

Ta có: \(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Theo PT₍₂₎: \(n_{BaCO_3}=n_{CO_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{BaCO_3}=0,2.197=39,4\left(g\right)\)

\(\Rightarrow\%_{m_{BaCO_3}}=\dfrac{39,4}{54,7}.100\%=72,03\%\)

\(\%_{m_{BaO}}=100\%-72,03\%=27,97\%\)

b. Ta có: \(m_{BaO}=54,7-39,4=15,3\left(g\right)\)

\(\Rightarrow n_{BaO}=\dfrac{15,3}{153}=0,1\left(mol\right)\)

\(\Rightarrow n_A=0,1+0,2=0,3\left(mol\right)\)

Theo PT_(1,2): \(n_{HCl}=2.n_A=2.0,3=0,6\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,6.36,5=21,9\left(g\right)\)

Ta có: \(C_{\%_{HCl}}=\dfrac{21,9}{m_{dd_{HCl}}}.100\%=20\%\)

\(\Rightarrow m_{dd_{HCl}}=109,5\left(g\right)\)

\(n_{CO2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

a) Pt : \(BaO+2HCl\rightarrow BaCl_2+H_2O|\)

             1            2               1           1

            0,1        0,2

            \(BaCO_3+2HCl\rightarrow BaCl_2+CO_2+H_2O|\)

                  1             2             1            1          1

                 0,2         0,4                         0,2

\(n_{BaCO3}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

\(m_{BaCO3}=0,2.197=39,4\left(g\right)\)

\(m_{BaO}=54,7-39,4=15,3\left(g\right)\)

⁰/₀BaO = \(\dfrac{15,3.100}{54,7}=27,97\)⁰/₀

⁰/₀BaCO₃ = \(\dfrac{39,4.100}{54,7}=72,03\)⁰/₀

b) Có : \(m_{BaO}=15,3\left(g\right)\)

\(n_{BaO}=\dfrac{15,3}{153}=0,1\left(mol\right)\)

\(n_{HCl\left(tổng\right)}=0,2+0,4=0,6\left(mol\right)\)

\(m_{HCl}=0,6.36,5=21,9\left(g\right)\)

\(m_{ddHCl}=\dfrac{21,9.100}{20}=109,5\left(g\right)\)

 Chúc bạn học tốt

             nCO2= 4,48 / 22.4= 0,2 ( mol)

a, BaO + 2HCl -> BaCl2 + H2O      (1)

BaCO3 + 2HCl-> BaCl2 + H2O + CO2   (2)

0,2                                                  <- 0,2       (mol)

Theo pt (2) ta có nBaCO3= 0,2 => mBaCO3= 0,2 x 197= 39,4 (g)

=> mBaO= 54,7- 39,4= 15.3 (g)

->% BaCO3= ( 39,4 : 54,7)x 100%= 72%

->% BaO= 100%-72% = 28%