Theo gt ta có: $n_{Al}=0,1(mol)$

a, $2Al+6HCl\rightarrow 2AlCl_3+3H_2$

b, $\Rightarrow n_{H_2}=0,15(mol)\Rightarrow V_{H_2}=3,36(l)$

c, Ta có: $n_{HCl}=0,3(mol)\Rightarrow m_{HCl}=10,95(g)\Rightarrow \%m_{ddHCl}=219(g)$

d, Bảo toàn khối lượng ta có: $m_{dd}=221,4(g)$

$\Rightarrow \%C_{AlCl_3}=6,02\%$

\(a) Fe + 2HCl \to FeCl_2\\ b) n_{HCl} = \dfrac{182,5.5\%}{36,5} = 0,25(mol)\\ n_{FeCl_2} = n_{H_2} = n_{Fe} = \dfrac{1}{2}n_{HCl} = 0,125(mol)\\ \Rightarrow m_{Fe} = 0,125.56 = 7(gam) ; V = 0,125.22,4 = 2,8(lít)\\ c) m_{dd\ sau\ phản\ ứng} = m_{Fe} + m_{dd\ HCl} - m_{H_2} = 7 + 182,5 - 0,125.2 = 189,25(gam)\\ C\%_{FeCl_2} = \dfrac{0,125.127}{189,25}.100\% = 8,39\%\)

`a)PTHH`

    `Fe + 2HCl -> FeCl_2 + H_2`

`0,125`  `0,25`           `0,125`     `0,125`        `(mol)`

`n_[HCl]=[5/100 .182,5]/[36,5]=0,25(mol)`

`b)m_[Fe]=0,125.56=7(g)`

  `V_[H_2]=0,125.22,4=2,8(l)`

`c)m_[HCl]=0,25.36,5=9,125(g)`

  `m_[FeCl_2]=0,125.127=15,875(g)`

`d)C%_[FeCl_2]=[15,875]/[7+182,5-0,125.2] .100~~8,39%`

\(a,m_{HCl}=\dfrac{182,5.5}{100}=9,125\left(g\right)\\ \rightarrow n_{HCl}=\dfrac{9,125}{36,5}=0,25\left(mol\right)\)

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

          0,125<-0,25--->0,125---->0,125

\(b,\left\{{}\begin{matrix}m_{Fe}=0,125.56=7\left(g\right)\\V_{H_2}=0,125.22,4=2,8\left(l\right)\end{matrix}\right.\\ c,m_{muối}=0,125.127=15,875\left(g\right)\\ d,m_{dd}=7+182,5-0,125.2=189\left(g\right)\\ \rightarrow C\%_{FeCl_2}=\dfrac{15,875}{189}.100\%=8,4\%\)

Fe+2HCl->FeCl₂+H₂

0,125--0,25---0,125-0,125

m _HCl=9,125 g=>n _HCl=\(\dfrac{9,125}{26,5}\)=0,25 mol

=>m Fe=0,125.56=7g

=>V_H2=0,125.22,4=2,8l

=>C%_FeCl2=\(\dfrac{0,125.127}{7+182,5-0,25}\).100=8,388%

HD:

a,

2AL+3H2SO4=>AL2(SO4)3+3H2

b,

Ta có: nAL=10.8/27=0.4(mol)

theo phương trình ta có: nH2=3/2nAL=0.6(mol)

=> VCO2=0.6*22.4=13.44(lít)

c,

Ta có: nH2=11.2/22.4=0.5(mol)

theo phương trình ta có: nH2SO4=nH2=0.5(mol)

=>mH2SO4=0.5*98=49(g)

ⁿH₂ = 6,72 : 22,4 = 0,3 ( mol ) 

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

                0,2     0,6 mol <-    0,3mol

a = ^mAl = 0,2 x 27 = 5,4 (g)

^VHCl = 0,6 : 2 = 0,3 ( l ) = 300 ( ml )

\(\left(a\right)2Al+3H_2O\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ \left(b\right)n_{H_2}=\dfrac{13,44}{22,4}=0,6mol\\ n_{Al}=\dfrac{0,6.2}{3}=0,4mol\\ m_{Al}=0,4.27=10,8g\\ \left(c\right)n_{O_2}=\dfrac{4,8}{32}=0,15mol\\ 4Al+3O_2\underrightarrow{t^0}2Al_2O_3\\ \Rightarrow\dfrac{0,4}{4}>\dfrac{0,15}{3}\Rightarrow Al.dư\\ n_{Al_2O_3}=\dfrac{0,15.2}{3}=0,1mol\\ m_{oxit}=m_{Al_2O_3}=0,1.102=10,2g\)

a: \(2Al+3H_2SO_4\rightarrow1Al_2\left(SO_4\right)_3+3H_2\uparrow\)

    0,4           0,6                0,2             0,6

b: \(n_{H_2}=\dfrac{13.44}{22.4}=0.6\left(mol\right)\)

=>\(n_{Al}=0.4\left(mol\right)\)

\(m_{Al}=0.4\cdot27=10.8\left(g\right)\)

c: \(4Al+3O_2\rightarrow2Al_2O_3\)

0,4                      0,2

\(m_{Al_2O_3}=0.2\left(27\cdot2+16\cdot3\right)=0.2\cdot102=20.4\left(g\right)\)

\(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

PTHH ;

2Al + 3H2SO4 --> Al2(SO4)3 + 3H2↑

  0,4           0,6               0,2           0,6

4Al + 3O2 ---> 2Al2O3

0,2       0,15          0,1 

\(n_{O_2}=\dfrac{4,8}{32}=0,15\left(mol\right)\)

\(\dfrac{0,4}{4}>\dfrac{0,15}{3}\)

--> Tính theo oxi

\(b,m_{Al}=0,4.27=10,8\left(g\right)\)

\(c,m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)

a/ PTHH

\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

b/

Ta có: \(n_{Al_2O_3}=\dfrac{10.2}{102}=0.1\left(mol\right)\)

\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

4                   2

x                    0.1

\(=>x=\dfrac{0.1\cdot4}{2}=0.2=n_{Al}\)  

\(=>m_{Al}=0.2\cdot27=5.4\left(g\right)\)

\(n_{H2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

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

             2            3                  1               3

             \(\dfrac{2}{15}\)                                             0,2

b) \(n_{Al}=\dfrac{0,2.2}{3}=\dfrac{2}{15}\left(mol\right)\)

⇒ \(m_{Al}=\dfrac{2}{15}.27=3,6\left(g\right)\)

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