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

\(n_{HCl}=\dfrac{14,6}{36,5}=0,4mol\)

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\)

a)Theo pt: \(n_{Fe}=n_{H_2}=0,1mol\)

\(\Rightarrow m_{Fe}=0,1\cdot56=5,6g\)

\(\%m_{Fe}=\dfrac{5,6}{12}\cdot100\%=46,67\%\)

\(\%m_{Cu}=100\%-46,67\%=53,33\%\)

c)\(m_{Cu}=12-5,6=6,4g\Rightarrow n_{Cu}=0,1mol\)

\(BTe:n_{O_2}=n_{Fe}+n_{Cu}=0,2mol\)

\(H=80\%\Rightarrow n_{O_2}=80\%\cdot0,2=0,16mol\)

\(n_{KMnO_4}=\dfrac{1}{2}n_{O_2}=0,08mol\)

\(\Rightarrow m_{KMnO_4}=0,08\cdot158=12,64g\)

\(V_{H_2}\)= \(\dfrac{V}{22,4}=\dfrac{6,72}{22,4}=0,3mol\)PTHH: 2Al+6HCL→2AlCl₃+3H2a)Theo pt: \(n_{Al_{ }}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}.0,3=0,2mol\)⇒ m_Al=n.M=0,2.27=5,4gb)Theo pt:\(n_{HCl}=\dfrac{6}{2}n_{H_2}=\dfrac{6}{2}.0,3=0,9mol\)⇒m_HCl=n.M=0,9.36.5=32,85g

\(n_{Fe}=\dfrac{5,6}{56}=0,1mol\)

\(n_{HCl}=\dfrac{14,6}{36,5}=0,4mol\)

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

0,1  <   0,4                           ( mol )

 0,1       0,2          0,1       0,1    ( mol )

Chất dư là HCl

\(m_{HCl\left(dư\right)}=\left(0,4-0,2\right).36,5=7,3g\)

\(V_{H_2}=0,1.22,4=2,24l\)

\(m_{FeCl_2}=0,1.127=12,7g\)

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\\ pthh:Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
\(LTL:\dfrac{0,1}{1}< \dfrac{0,4}{1}\) 
=> H₂SO₄ d 
\(n_{H_2SO_4\left(pu\right)}=n_{Fe}=0,1\left(mol\right)\\ m_{H_2SO_4\left(d\right)}=\left(0,4-0,1\right).98=29,4g\) 
\(n_{H_2}=n_{FeSO_4}=n_{Fe}=0,1\left(mol\right)\)
\(V_{H_2}=0,1.22,4=2,24l\\ m_{FeSO_4}=0,1.152=15,2g\)

a)

$Fe + 2HCl \to FeCl_2 + H_2$
n H2 = n Fe = 11,2/56 = 0,2(mol)

V H2 = 0,2.22,4 = 4,48(lít)

b)

n HCl = 2n Fe = 0,2.2 = 0,4(mol)

=> CM HCl = 0,4/0,4 = 1M

c)

$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$

Ta thấy :

n CuO = 64/80 = 0,8 > n H2 = 0,2 nên CuO dư

Theo PTHH :

n CuO pư = n Cu = n H2 = 0,2(mol)

n Cu dư = 0,8 - 0,2 = 0,6(mol)

Vậy : 

%m Cu = 0,2.64/(0,2.64 + 0,6.80)   .100% = 21,05%
%m CuO = 100% -21,05% = 78,95%

zn+2Hcl→zncl2+h2

a)        nZn=6.5/65=0.1(mol)

         ta có nZn/1=0.1/1<nHcl/2=0.25/2=0.125

           →zn hết ,hcl dư

         theo pt:nH2=nZn=0.1 (mol)

     Vh2=0.1*22.4=2.24

b)    sau pư zn hết ,Hcl dư 

theo pt nHcl=2nZn=0.1*2=0.2( mol )

sô mol Hcl dư là:

0.25-0.2=0.05(mol)

mHcl dư là:0.05*36.5=1.825

xong

a)

\(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)

\(n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

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

PTHH: Fe + 2HCl --> FeCl₂ + H₂

                      0,5----------->0,25

=> V_H2 = 0,25.22,4 = 5,6 (l)

b)\(n_{Fe_3O_4}=\dfrac{13,92}{232}=0,06\left(mol\right)\)

PTHH: Fe₃O₄ + 4H₂ --t^o--> 3Fe + 4H₂O

Xét tỉ lệ: \(\dfrac{0,06}{1}< \dfrac{0,25}{4}\) => Fe₃O₄ hết, H₂ dư

PTHH: Fe₃O₄ + 4H₂ --t^o--> 3Fe + 4H₂O

             0,06-->0,24------->0,18-->0,24

=> \(\left\{{}\begin{matrix}m_{Fe}=0,18.56=10,08\left(g\right)\\m_{H_2O}=0,24.18=4,32\left(g\right)\\m_{H_2\left(dư\right)}=\left(0,25-0,24\right).2=0,02\left(g\right)\end{matrix}\right.\)

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,\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right);n_{H_2SO_4}=\dfrac{29,4}{98}=0,3\left(mol\right)\)

PTHH: 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

Mol:    0,1         0,15            0,05             0,15

b,Ta có: \(\dfrac{0,1}{2}< \dfrac{0,3}{3}\) ⇒ Al hết, H₂SO₄ dư

\(\Rightarrow m_{H_2SO_4dư}=\left(0,3-0,15\right).98=14,7\left(g\right)\)

c, \(m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(g\right)\)

d, \(V_{H_2}=0,15.22,4=3,36\left(l\right)\)