\(M_{H_2}=1.2=2g/mol\\ M_{O_2}=16.2=32g/mol\\ M_{CO_2}=12+16.2=44g/mol\\ M_{Al_2O_3}=27.2+16.3=102g/mol\\ M_{Fe}=56g/mol\\ M_{Mg}=24g/mol\)

a) Mg + 2HCl --> MgCl₂ + H₂

b) \(n_{Mg}=\dfrac{9,6}{24}=0,4\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,4}{1}>\dfrac{0,5}{2}\) => Mg dư,HCl hết

PTHH: Mg + 2HCl --> MgCl₂ + H₂

          0,25<--0,5--->0,25--->0,25

=> n_Mg(dư) = 0,4 - 0,25 = 0,15 (mol)

c) \(V_{H_2}=0,25.22,4=5,6\left(l\right)\)

\(m_{MgCl_2}=0,25.95=23,75\left(g\right)\)

Gọi \(n_{Fe}=a\left(mol\right)\rightarrow n_{Mg}=\dfrac{1}{1}.a=a\left(mol\right)\)

\(\rightarrow n_{Zn}=0,3-a-a=0,3-2a\left(mol\right)\)

\(\rightarrow65\left(0,3-2a\right)+56a+24a=13\\ \Leftrightarrow a=0,13\\ \Rightarrow\left\{{}\begin{matrix}n_{Fe}=n_{Mg}=0,13\left(mol\right)\\n_{Zn}=0,3-0,13.2=0,04\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{56.0,13}{13}.100\%=56\%\\\%m_{Mg}=\dfrac{24.0,13}{13}.100\%=24\%\\\%m_{Zn}=100\%-56\%-25\%=20\%\end{matrix}\right.\)

PTHH:

Fe + 2HCl ---> FeCl₂ + H₂

Zn + 2HCl ---> ZnCl₂ + H₂

Mg + 2HCl ---> MgCl₂ + H₂

Theo pthh: n_H2 = n_{kim loại} = 0,3 (mol)

\(n_{CuO}=\dfrac{80}{80}=1\left(mol\right)\)

PTHH: CuO + H₂ --t^o--> Cu + H₂O

LTL: 1 > 0,3 => CuO dư

Chất rắn sau pư gồm: CuO dư, Cu

Theo pthh: n_{CuO (pư)} = n_Cu = n_H2 = 0,3 (mol)

=> m_{chất rắn} = 80,(1 - 0,3) + 64.0,3 = 75,2 (g)

Mg+2HCl->MgCl2+H2

0,15--0,3--------------0,15

CuO+2HCl->CuCl2+H2O

0,1------0,2

n H2=\(\dfrac{3,36}{22,4}\)=0,15 mol
=>%m Mg=\(\dfrac{0,15.24}{11,6}.100=31,03\%\)

=>m CuO=8g =>n CuO=\(\dfrac{8}{80}\)=0,1 mol

=>%m CuO=68,97%

=>CM HCl=\(\dfrac{0,3+0,2}{0,2}\)=2,5M

\(a.\)

\(m_{hh}=0.12\cdot90+0.15\cdot58=19.5\left(g\right)\)

\(b.\)

\(V_{hh}=\left(0.25+0.1+0.05\right)\cdot22.4=8.96\left(l\right)\)

\(c.\)

\(n_A=\dfrac{10.08}{22.4}=0.45\left(mol\right)\)

\(M_A=23\cdot2=46\left(\dfrac{g}{mol}\right)\)

\(m_A=0.45\cdot46=20.7\left(g\right)\)

\(d.\)

\(n_{hh}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)

Vì CO₂ : O₂ = 2 : 1 

\(\Rightarrow n_{CO_2}=0.2\left(mol\right),n_{O_2}=0.1\left(mol\right)\)

\(m_{hh}=0.2\cdot44+0.1\cdot32=12\left(g\right)\)

\(\overline{M}=\dfrac{12}{0.3}=40\left(\dfrac{g}{mol}\right)\)

a) Mg + 2HCl --> MgCl₂ + H₂

b) Xét tỉ lệ: \(\dfrac{0,15}{1}>\dfrac{0,2}{2}\) => HCl hết, Mg dư

PTHH: Mg + 2HCl --> MgCl₂ + H₂

            0,1<--0,2-------------->0,1

=> V_H2 = 0,1.22,4 = 2,24 (l)

c) m_Mg(dư) = 24(0,15 - 0,1) = 1,2 (g)

a) Mg (0,1 mol) + 2HCl (0,2 mol) \(\rightarrow\) MgCl₂ + H₂\(\uparrow\) (0,1 mol).

b) Thể tích khí hiđro thu được là 0,1.22,4=2,24 (lít).

c) Khối lượng magie dư sau phản ứng là (0,15-0,1).24=1,2 (g).

a

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

0,15<-0,15<--0,15<----0,15

\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

0,16-->0,32---->0,16

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

\(m_{Mg}=0,15.24=3,6\left(g\right)\\ m_{MgO}=10-3,6=6,4\left(g\right)\)

b

\(\%m_{Mg}=\dfrac{3,6.100\%}{10}=36\%\\ \%m_{MgO}=\dfrac{6,4.100\%}{10}=64\%\)

c

\(n_{MgO}=\dfrac{6,4}{40}=0,16\left(mol\right)\)

\(CM_{HCl}=\dfrac{0,15+0,32}{0,2}=2,35M\)

d

\(m_{MgCl_2}=\left(0,15+0,16\right).95=29,45\left(g\right)\)

e

\(CM_{MgCl_2}=\dfrac{0,15+0,16}{0,2}=1,55M\)

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

Phương trình hóa học

Mg + 2HCl ---> MgCl₂ + H₂ (**)

MgO + 2HCl ---> MgCl₂ + H₂O (***)

b) Từ (*) và (**) ta có \(n_{Mg}=0,15\Leftrightarrow m_{Mg}=0,15.24=3,6\left(g\right)\)

\(\Rightarrow m_{MgO}=10-3,6=6,4\left(g\right)\)

 \(\%Mg=\dfrac{3,6}{10}.100\%=36\%\)

\(\%MgO=\dfrac{6,4}{10}.100\%=64\%\)

c) Xét phản ứng (**) ta có \(m_{MgO}=6,4\left(g\right)\Leftrightarrow n_{MgO}=n_{MgCl_2}=\dfrac{1}{2}n_{HCl}=0,16\left(mol\right)\) (1) 

\(\Leftrightarrow n_{HCl}=0,32\left(mol\right)\)

Tương tự có số mol HCl trong phản ứng (*) là 0,3 mol

\(C_M=\dfrac{0,32+0,3}{0,2}=3,1\left(M\right)\)

d) Từ (1) ; (*) ; (**) ta có : \(n_{MgCl_2}=0,15+0,16=0,31\left(mol\right)\)

\(m_{MgCl_2}=0,31.95=29,45\left(g\right)\)

e) \(C_M=\dfrac{0,31}{0,2}=1,55\left(M\right)\)