1/ PT : X +  2H₂O -> X[OH]₂ + H₂

_{mol :    \(\frac{6}{M_X}\)}             ->                   \(\frac{6}{M_X}\)     

=> m_H2 = \(\frac{12}{M_X}\)       => m_dd = m+6 - \(\frac{12}{M_X}\)

Ta có: m+5,7 = m+6 - \(\frac{12}{M_X}\)

<=> \(\frac{12}{M_X}\)= 0,3 => M_{X = 40 => X là Canxi [Ca]}

2/ Dặt n_HCl= a [a> 0] => m_ddHCl= 36,5a : 14,6 x 100= 250a

     PT : X  + 2HCL => XCl₂ +  H₂

   mol :   a/2        a         ->  a/2        a/2

m_H2 = a/2 x 2 = a ; m_X = a/2 . M_X

m XCl₂= a/2 x [M_X +71]

m_dd XCL2= a/2 .M_X + 250a  - a = a/2 .M_X +249a

Ta có :\(\frac{\frac{a}{2}\times M_X+\frac{71}{2}a}{M_X\times a:2+249a}\times100\%=24,15\%\)

<=> \(\frac{M_X+71}{M_X+498}=24,15\%\Leftrightarrow M_X=65\)=> X là kẽm [Zn]

\(n_{H_2}=\dfrac{50.4}{22.4}=2.25\left(mol\right)\)

\(2M+2nH_2SO_4\rightarrow M_2\left(SO_4\right)_n+nH_2\)

\(\dfrac{4.5}{n}.............................2.25\)

\(M_M=\dfrac{54}{\dfrac{4.5}{n}}=12n\)

\(BL:n=2\Rightarrow M=24\)

\(M:Mg\)

\(n_{H_2SO_4}=n_{H_2}=2.25\left(g\right)\)

\(m_{dd_{H_2SO_4}}=\dfrac{2.25\cdot98\cdot100}{10}=2205\left(g\right)\)

Giả sử kim loại cần tìm là A có hóa trị n không đổi.

Ta có: \(n_{H_2}=\dfrac{50,4}{22,4}=2,25\left(mol\right)\)

BT e, có: \(n_A=\dfrac{2,25.2}{n}=\dfrac{4,5}{n}\left(mol\right)\)

\(\Rightarrow M_A=\dfrac{54}{\dfrac{4,5}{n}}=12n\)

Với n = 1 ⇒ M_A = 12 (loại)

Với n = 2 ⇒ M_A = 24 (nhận)

Vậy: A là Magie (Mg).

BTNT H có: \(n_{H_2SO_4}=n_{H_2}=2,25\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=2,25.98=220,5\left(g\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{220,5.100}{10}=2205\left(g\right)\)

Bạn tham khảo nhé!

\(n_{H_2SO_4}=\dfrac{78,4.20}{100.98}=0,16\left(mol\right)\)

PTHH: X + H₂SO₄ --> XSO₄ + H₂

____0,16<--0,16--->0,16-->0,16

=> 0,16.M_X + 78,4 - 0,16.2 = 88,48

=> M_X = 65 (g/mol)

=> X là Zn

_{Không viết phương trình nhá !!}

a) Gọi a và b lần lượt là số mol của Mg và Al

\(\Rightarrow24a+27b=1,035\)  (1)

Ta có: \(n_{H_2}=\dfrac{1,176}{22,4}=0,0525\left(mol\right)\)

Bảo toàn electron: \(2a+3b=2\cdot0,0525\)  (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,015\\b=0,025\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,015\cdot24}{1,035}\cdot100\%\approx34,78\%\\\%m_{Al}=65,22\%\end{matrix}\right.\)

b) Ta có: \(\left\{{}\begin{matrix}\Sigma n_{H_2SO_4}=\dfrac{100\cdot9,8\%}{98}=0,1\left(mol\right)\\n_{H_2SO_4\left(p/ứ\right)}=n_{H_2}=0,0525\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,0475\left(mol\right)\) \(\Rightarrow m_{H_2SO_4\left(dư\right)}=0,0475\cdot98=4,655\left(g\right)\)

c) Bảo toàn nguyên tố: \(\left\{{}\begin{matrix}n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,0125\left(mol\right)\\n_{MgO}=n_{Mg}=0,015\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow m_{oxit}=0,0125\cdot102+0,015\cdot40=1,875\left(g\right)\)

a) Gọi \(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\)

\(m_A=1,035\left(g\right)\rightarrow24a+27b=1,035\) (1)

\(Mg+2H_2SO_4đ\rightarrow MgSO_4+SO_2+2H_2O\)

a ------------ 2a ----------------------- a (mol) 

\(2Al+6H_2SO_4đ\rightarrow Al_2\left(SO_4\right)_3+3SO_2+6H_2O\)

b ------------ 3b -------------------------- 1,5b (mol)

\(n_{SO_2}=\dfrac{1,176}{22,4}=0,0525\left(mol\right)\rightarrow a+1,5b=0,0525\) (2) 

Giải hệ (1)(2) \(\rightarrow\left\{{}\begin{matrix}a=0,015\\b=0,025\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}m_{Mg}=0,015.24=0,36\left(g\right)\\m_{Al}=0,025.27=0,675\left(g\right)\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}\%m_{Mg}=34,78\%\\\%m_{Al}=65,22\%\end{matrix}\right.\)

b) \(\Sigma_{n_{H_2SO_4}}=2a+3b=0,105\left(mol\right)\)

\(\rightarrow m_{H_2SO_4}=0,105.98=10,29\left(g\right)\)

c. \(\left\{{}\begin{matrix}n_{MgO}=n_{Mg}=0,015\left(mol\right)\\n_{Al_2O_3}=\dfrac{1}{2}.n_{Al}=0,0125\left(mol\right)\end{matrix}\right.\)

\(\rightarrow m_{oxit}=0,015.40+0,0125.102=1,875\left(g\right)\)

\(Mg+2H_2SO_{4\left(đ,n\right)}\)\(\rightarrow MgSO_4+SO_2+2H_2O\)

\(Zn+2H_2SO_4\rightarrow ZnSO_4+SO_2+2H_2O\)

Đặt \(n_{Mg}=x\left(mol\right);n_{Zn}=y\left(mol\right)\)

Có hệ:\(\left\{{}\begin{matrix}24x+65y=6,85\\120x+161y=52,1\end{matrix}\right.\) \(\Rightarrow y=-\dfrac{357}{3280}\) (sai đề ?)

nHCl = 200.0,073/36,5 = 0,4(mol)
2M + 2nHCl → 2MCln + nH2
0,4/n _ 0,4 ___ 0,4/n __ 0,2

mddMCln = 200 + M.0,4/n - 0,2.2 = 199,6 + 0,4M/n
Nồng độ muối MCln bằng 11,96%
⇒ (M + 35,5n).0,4/n = 0,1196.(199,6 + 0,4M/n)
⇒ M = 27,5n
n = 2, M = 55 (Mn)
x = 55.0,4/2 = 11(g)

nH2 = 1,344 : 22,4 = 0,06(mol) 
pthh 2M+ 3H2SO4 --->  M2(SO4)3+ 3H2 
       0,04<--  0,06---------------------------0,06(mol) 
M _M = 1,08 : 0,04 = 27 (g/mol ) 
=> M : Al 
mH2SO4 = 0,06.98  =5,88 (g)

nH2 = 1,344/22,4 = 0,06 (mol)

PTHH: 

2M + 3H2SO4 -> Al2(SO4)3 + 3H2

0,04 <--- 0,06 <--- 0,02 <--- 0,06

M(M) = 1,08/0,04 = 27 (g/mol(

=> M là Al

mH2SO4 = 0,06 . 98 = 5,88 (g)

a, m_{chất rắn} = m_Cu = 12,8 (g)

=> m_{hh (Al, Zn)} = 28,5 - 12,8 = 16,7 (g)

\(m_{H_2SO_4}=7,84\%.500=39,2\left(g\right)\\ n_{H_2SO_4}=\dfrac{39,2}{98}=0,4\left(mol\right)\)

Gọi \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Zn}=b\left(mol\right)\end{matrix}\right.\left(a,b>0\right)\)

PTHH:

2Al + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂↑

a----->1,5a---------->0,5a-------->1,5a

Zn + H₂SO₄ ---> ZnSO₄ + H₂

b---->b------------>b--------->b

m_{dd (tăng)} = m_{hh (Al, Zn)} - m_H2 = 27a + 65a - 2.(1,5a - b) = 24a - 63b = 515 - 500 = 15 (g)

=> Hệ pt \(\left\{{}\begin{matrix}27a+65b=15,7\\24a-63b=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,2\left(mol\right)\end{matrix}\right.\left(TM\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{12,8}{28,5}.100\%=44,9\%\\\%m_{Al}=\dfrac{0,1.27}{28,5}.100\%=18,9\%\\\%m_{Zn}=100\%-44,9\%-18,9\%=36,2\%\end{matrix}\right.\)

b, \(n_{H_2SO_{4\left(dư\right)}}=0,4-0,1.1,5-0,2=0,05\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0,05.342}{515}.100\%=3,32\%\\C\%_{ZnSO_4}=\dfrac{0,2.161}{515}.100\%=6,25\%\\C\%_{H_2SO_{4\left(dư\right)}}=\dfrac{0,05.98}{515}.100\%=0,95\%\end{matrix}\right.\)