=>AO+2HCL->ACL2+H2O(1)

=>HCL+NaOH->NaCL+H2O(2)

(2)=>\(nHCL=nNaOH=\dfrac{100}{1000}.0,1=0,01mol\)

\(\)\(=>nHCL\left(1\right)=\dfrac{500}{1000}.0,1-0,01=0,04mol\)

\(=>nAO=\dfrac{1}{2}nHCL=>=\dfrac{0,8}{A+16}=0,02=>A=24g/mol\)

=>A là Mg =>ct oxit : MgO

a) CT oxit \(AO\)

\(AO+2HCl\rightarrow ACl_2+H_2\\ n_{HCl}=0,4\left(mol\right)\\ n_A=\dfrac{1}{2}n_{HCl}=0,2\left(mol\right)\\ \Rightarrow M_{AO}=A+16=\dfrac{8}{0,2}=40\\ \Rightarrow A=24\left(Mg\right)\)

b)\(n_{MgSO_3}=\dfrac{10,4}{104}=0,1\left(mol\right)\\ n_{H_2SO_4}=\dfrac{200.24,5\%}{98}=0,5\left(mol\right)\\ MgSO_3+H_2SO_4\rightarrow MgSO_4+SO_2+H_2O\\ LTL:\dfrac{0,1}{1}< \dfrac{0,5}{1}\\ \Rightarrow H_2SO_4dưsauphảnứng\\ n_{H_2SO_4\left(pứ\right)}=n_{SO_2}=n_{MgSO_4}=n_{MgSO_3}=0,1\left(mol\right)\\ n_{H_2SO_4\left(dư\right)}=0,5-0,1=0,4\left(mol\right)\\ m_{ddsaupu}=10,4+200-0,1.64=204\left(g\right)\\ C\%_{MgSO_4}=\dfrac{0,1.120}{204}.100=5,88\%\\ C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,4.98}{204}=19,22\%\)

\(a,n_{AO}=\dfrac{8}{M_A+16}(mol);n_{HCl}=1.0,4=0,4(mol)\\ PTHH:AO+2HCl\to ACl_2+H_2O\\ \Rightarrow n_{AO}=\dfrac{1}{2}n_{HCl}=0,2(mol)\\ \Rightarrow M_{AO}=\dfrac{8}{0,2}=40(g/mol)\\ \Rightarrow M_{A}=40-16=24(g/mol)\\ \text {Vậy A là magie(Mg) và CTHH oxit là }MgO\\\)

\(b,n_{MgSO_3}=\dfrac{10,4}{104}=0,1(mol)\\ m_{H_2SO_4}=\dfrac{200.24,5\%}{100\%}=49(g)\\ \Rightarrow n_{H_2SO_4}=\dfrac{49}{98}=0,5(mol)\\ PTHH:MgSO_3+H_2SO_4\to MgSO_4+SO_2\uparrow +H_2O \)

Vì \(\dfrac{n_{MgSO_3}}{1}<\dfrac{n_{H_2SO_4}}{1}\) nên \(H_2SO_4\) dư

\(\Rightarrow n_{MgSO_4}=n_{SO_2}=n_{H_2O}=n_{MgSO_3}=0,1(mol)\\ \Rightarrow \begin{cases} m_{CT_{MgSO_4}}=0,1.120=12(g)\\ m_{SO_2}=0,1.64=6,4(g)\\ m_{H_2O}=0,1.18=1,8(g) \end{cases}\\ \Rightarrow m_{dd_{MgSO_4}}=10,4+200-6,4-1,8=202,2(g)\\ \Rightarrow C\%_{MgSO_4}=\dfrac{12}{202,2}.100\%\approx 5,93\%\)

Bài 24:

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ PTHH:A+2HCl\rightarrow ACl_2+H_2\uparrow\)

Theo pthh: n_A = n_H2 = 0,15 (mol)

=> M_A = \(\dfrac{3,6}{0,15}=24\left(\dfrac{g}{mol}\right)\)

=> A là Mg

Bài 25:

\(n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\\ PTHH:2A+6HCl\rightarrow2ACl_3+3H_2\uparrow\\ Mol:0,3\leftarrow0,9\leftarrow0,3\leftarrow0,45\\ \rightarrow\left\{{}\begin{matrix}M_A=\dfrac{8,1}{0,3}=27\left(\dfrac{g}{mol}\right)\Rightarrow A:Al\\m_{HCl}=0,9.36,5=32,85\left(g\right)\end{matrix}\right.\)

Bài 24.

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\)

\(n_A=\dfrac{3,6}{M_A}\) mol

\(A+2HCl\rightarrow ACl_2+H_2\)

0,15                           0,15   ( mol )

\(\Rightarrow\dfrac{3,6}{M_A}=0,15mol\)

\(\Leftrightarrow M_A=24\) ( g/mol )

=> A là Magie ( Mg )

Bài 25.

\(n_{H_2}=\dfrac{10,08}{22,4}=0,45mol\)

\(n_A=\dfrac{8,1}{M_A}\) mol

\(2A+6HCl\rightarrow2ACl_3+3H_2\)

0,3                                0,45  ( mol )

\(\Rightarrow\dfrac{8,1}{M_A}=0,3\)

\(\Leftrightarrow M_A=27\) g/mol

=> A là nhôm ( Al )

a,Gọi hỗn hợp 2 kim loại là R

n_H2 =\(\dfrac{3,36}{22,4}\)= 0,15 mol

PTHH: R + HCL\(\rightarrow\) RCL+\(\dfrac{1}{2}\)H2

TBR: 0,3 \(\leftarrow\)0,15

M_R= \(\dfrac{8,5}{0,3}\)=28,3

\(\Rightarrow\) hai kim loại đó là Na và K

b,

a) Gọi KL cần tìm là X 
n_HCl=\(\frac{5,6}{22,4}\)=0,25 
PTHH: X + HCl \(\rightarrow\) XCl₂ + H₂
           0,25 0,5      0,25 0,25 
\(\Rightarrow\)m_X = \(\frac{16.25}{0,25}\)=65g ( Zn ) 
b) mHCl= \(0,5.36,5\)=18.25g 
mdd= \(\frac{18.25}{0,1825}\)=100g 
Cm = \(\frac{0,5}{\frac{0,1}{0,2}}\)=6 mol/l 
c) C% = 0,25.(65+71)/(100+16,25-0,5).100=29.73% 

a) Gọi kl cần tìm là X
nHCl= 5.6/22.4=0.25
PTHH: X + HCl -> XCl2 + H2
0.25 0.5 0.25 0.25
=>mX = 16.25/0.25=65g ( Zn )
b) mHCl= 0.5*36.5=18.25g
mdd= 18.25/0.1825=100g
Cm = 0.5/(0.1/1.2)=6 mol/l (lơn z tar)
c) C% = 0.25*(65+71)/(100+16.25-0.5)*100=29.73%

a)
Fe +2 HCl --> FeCl2+ H2
R + 2HCl --> RCl2 + H2
nhh = nH2= 0,672/22,3=0,03 mol
=> M trung bình =1,52/ 0,03= 50,6
Vì MFe=56> Mtb => R<50,6
nH2SO4= 49.8%/98= 0,04 mol
R + H2SO4 --> RSO4 +H2
axit dư => nR phản ứng = nH2SO4 phản ứng <0,04
=> R>1,52/0,04= 38
Suy ra 38< R< 50,6, R hóa trị II
Vậy R là Canxi (Ca) . R =40
b)
Có nFe +nCa= 0,03mol
m hh=56nFe +40nCa =1,52
=> n Fe = 0,02 mol ; n Ca = 0,01 mol
=> %mFe, %mCa
c) 
nHCl = 2nH2 =0,06 mol
=> mddHCl =36,5.0,06.100/15=14,6g
mddB= m hhkl + m ddHCl - mH2 = 1,52+ 14,6 -0,03.2= 16,06 g
dd B có FeCl2 0,02 mol và CaCl2 0,01 mol
​=> C%.