a)

$2Al +3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
b)

$n_{H_2} = n_{H_2SO_4} = \dfrac{300.9,8\%}{98} = 0,3(mol)$
$V_{H_2} = 0,3.22,4 = 6,72(lít)$

c)

$n_{Al_2(SO_4)_3} = \dfrac{1}{3}n_{H_2SO_4} = 0,1(mol)$
$m_{Al_2(SO_4)_3} = 0,1.342 = 34,2(gam)$

d)

$n_{Al} = \dfrac{2}{3}n_{H_2SO_4} = 0,2(mol)$
$m_{dd} = 0,2.27 + 300 - 0,3.2 = 304,8(gam)$
$C\%_{Al_2(SO_4)_3} = \dfrac{34,2}{304,8}.100\% = 11,22\%$

nH2SO4=0,3(mol)

PTHH: 2Al + 3 H2SO4 -> Al2(SO4)3 + 3 H2

a) 0,2_______0,3______0,1______0,3(mol)

b) V(H2,đktc)=0,3.22,4=6,72(l)

c) a=mAl=0,2.27=5,4(g)

=>a=5,4(g)

d)  mAl2(SO4)3=342.0,1=34,2(g)

e) mddAl2(SO4)3= 5,4+ 300 - 0,3.2= 304,8(g)

=>C%ddAl2(SO4)3= (34,2/304,8).100=11,22%

a) Phương trình hóa học:

              2Al+ 3H₂SO₄→ Al₂(SO₄)₃+ 3H₂

( mol)      0,2       0,3            0,1          0,3                                        

b) m H₂SO₄= \(\dfrac{9,8\%.300}{100\%}=29,4\)(gam)

→n H₂SO₄= \(\dfrac{m}{M}=\dfrac{29,4}{98}=0,3\left(mol\right)\)

=> m_Al= \(n.m=0,2.27=5,4\left(gam\right)\)

c) V H₂= n.22,4= 0,3.22,4= 6,72( lít)

m H₂= n.M= 0,3.2= 0,6(gam)

d) m Al₂(SO₄)₃= n.M= 0,1.342= 34,2(gam)

e) m_{dd sau phản ứng}= m_Al+ m_ddH₂SO₄- m H₂

                                    = 5,4  + 300- 0,6= 304,8(gam)

=> C%_{dd sau phản ứng}=\(\dfrac{34,2}{304,8}.100\%=11,22\%\)

a) 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

b) \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

            0,2<----0,3<-----------------0,3

=> m_Al = 0,2.27 = 5,4 (g)

c) \(m_{dd.H_2SO_4}=\dfrac{0,3.98}{30\%}=98\left(g\right)\)

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

\(n_{CuO}=\dfrac{20}{80}=0,25\left(mol\right)\)

PTHH: Zn + 2HCl ---> ZnCl₂ + H₂

          0,1---->0,1----------------->0,1

=> \(\left\{{}\begin{matrix}V_{H_2}=0,1.22,4=2,24\left(l\right)\\C_{M\left(HCl\right)}=\dfrac{0,2}{0,2}=1M\end{matrix}\right.\)

PTHH: CuO + H₂ --t^o--> Cu + H₂O
LTL: 0,25 > 0,1 => CuO dư

Theo pthh: n_Cu = n_H2 = 0,1 (mol)

=> m_Cu = 0,1.64 = 6,4 (g)

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\) 
         0,1    0,2                           0,2 
\(V_{H_2}=0,2.22,4=4,48l\\ C_M=\dfrac{0,2}{0,2}=1M\\ n_{CuO}=\dfrac{20}{80}=0,25\left(G\right)\\ pthh:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ LTL:0,25>0,1\) 
=>CuO dư 
\(n_{Cu}=n_{H_2}=0,1\left(mol\right)\\ m_{Cu}=0,1.64=6,4g\)

\(n_{Zn}=\dfrac{8,125}{65}=0,125\left(mol\right)\\ m_{HCl}=\dfrac{100.18,25}{100}=18,25\left(g\right)\\ n_{HCl}=\dfrac{18,25}{36,5}=0,5\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)  
            0,125                            0,125 (mol ) 
\(\Rightarrow V_{H_2}=0,125.22,4=2,8\left(l\right)\\ \)   
\(C\%=\dfrac{8,125}{8,125+18,25}.100\%=30,8\%\)
 

Bài 18:

Ta có: \(n_{Zn}=\dfrac{8,125}{65}=0,125\left(mol\right)\)

\(m_{HCl}=100.18,25\%=18,25\left(g\right)\Rightarrow n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)

a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, Xét tỉ lệ: \(\dfrac{0,125}{1}< \dfrac{0,5}{2}\), ta được HCl dư.

Theo PT: \(n_{H_2}=n_{Zn}=0,125\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,125.22,4=2,8\left(g\right)\)

\(m_{H_2}=0,125.2=0,25\left(g\right)\)

c, Theo PT: \(\left\{{}\begin{matrix}n_{ZnCl_2}=n_{Zn}=0,125\left(mol\right)\\n_{HCl\left(pư\right)}=2n_{Zn}=0,25\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{HCl\left(dư\right)}=0,25\left(mol\right)\)

Có: m _{dd sau pư} = 8,125 + 100 - 0,25 = 107,875 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,125.136}{107,875}.100\%\approx15,76\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,25.36,5}{107,875}.100\%\approx8,46\%\end{matrix}\right.\)

Bạn tham khảo nhé!

`a)PTHH:`

`Zn + 2HCl -> ZnCl_2 + H_2↑`

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

`b)n_[Zn] = [ 8,125 ] / 65 = 0,125 (mol)`

`n_[HCl] = [ [ 18,25 ] / 100 . 100 ] / [ 36,5 ] = 0,5 (mol)`

Ta có: `[ 0,125 ] / 1 < [ 0,5 ] / 2`

    `-> Zn` hết, `HCl` dư

 `=> V_[H_2] = 0,125 . 22,4 = 2,8 (l)`

 `=> m_[H_2] = 0,125 . 2 = 0,25 (g)`

`c)m_\text{dd sau p/ứ} = 8,125 + 100 - 0,25 = 107,875 (g)`

`=> C%_[ZnCl_2] = [ 0,125 . 136 ] / [ 107,875 ] . 100 ~~ 15,76%`

`=> C%_[HCl(dư)] = [ ( 0,5 - 0,25 ) . 36,5 ] / [ 107,875 ] . 100 ~~ 8,46%`

Ta có: \(n_{Al}=\dfrac{4,86}{27}=0,18\left(mol\right)\)

PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

a, \(n_{HCl}=3n_{Al}=0,54\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,54}{0,15}=3,6\left(M\right)\)

b, \(n_{AlCl_3}=n_{Al}=0,18\left(mol\right)\Rightarrow m_{AlCl_3}=0,18.133,5=24,03\left(g\right)\)

c, \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,27\left(mol\right)\Rightarrow V_{H_2}=0,27.24,79=6,6933\left(l\right)\)

 PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)

Làm gộp các phần còn lại

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Al_2\left(SO_4\right)_3}=0,1mol\\n_{H_2SO_4}=n_{H_2}=0,3mol\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\m_{Al_2\left(SO_4\right)_3}=0,1\cdot342=34,2\left(g\right)\\m_{H_2SO_4}=0,3\cdot98=29,4\left(g\right)\end{matrix}\right.\) 

\(n_{Fe}=\dfrac{m}{M}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ PTHH:Fe+2HCl->FeCl_2+H_2\)

tỉ lệ:          1      :    2     :     1        :     1

n(mol)      0,2--->0,4------->0,2----->0,2

\(m_{FeCl_2}=n\cdot M=0,2\cdot\left(56+35,5\cdot2\right)=25,4\left(g\right)\\ V_{H_2\left(dktc\right)}=n\cdot22,4=0,2\cdot22,4=4,48\left(l\right)\\ V_{H_2\left(dkc\right)}=n\cdot24,79=4,958\left(l\right)\)

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

b) \(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

Theo phương trình hóa học: \(n_{H_2}=n_{Fe}=o,2\left(mol\right)\)

\(V_{H_2\left(dktc\right)}=n_{H_2}\times22,4=0,2\times22,4=4,48\left(l\right)\)

\(V_{H_2\left(dkc\right)}=n_{H_2}\times24,79=0,2\times24,79=4,96\left(l\right)\)

c) Theo phương trình hóa học: \(n_{FeCl_2}=n_{H_2}=0,2\left(mol\right)\)

\(m_{FeCl_2}=n_{FeCl_2}\times M_{FeCl_2}=0,2.127=25,4\left(g\right)\)

a, PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

b, Ta có: \(n_{Fe}=\dfrac{19,6}{56}=0,35\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Fe}=0,35\left(mol\right)\Rightarrow V_{H_2}=0,35.22,4=7,84\left(l\right)\)

c, \(n_{H_2SO_4}=n_{Fe}=0,35\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,35}{0,2}=1,75\left(M\right)\)

d, \(n_{FeSO_4}=n_{Fe}=0,35\left(mol\right)\Rightarrow m_{FeSO_4}=0,35.152=53,2\left(g\right)\)

e, \(C_{M_{FeSO_4}}=\dfrac{0,35}{0,2}=1,75\left(M\right)\)

d, \(n_{H_2SO_4}=0,25.1,6=0,4\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{n_{Fe}}{1}< \dfrac{n_{H_2SO_4}}{1}\), ta được H₂SO₄ dư.

Theo PT: \(n_{H_2SO_4\left(pư\right)}=n_{Fe}=0,35\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,4-0,35=0,05\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4\left(dư\right)}=0,05.98=4,9\left(g\right)\)

2Al+6HCl-->2AlCl₃+3H₂

0,3----0,9---------0,3------0,45

=>n _Al=8,1\17=0,3 mol

=>V_H2=0,45.22,4=10,08l

=>m _HCl=0,9.26,5=32,85g

=>m_AlCl3=0,3.133,5=40,05g

C2 :Bảo Toàn khối lượng 

=>m _AlCl3=40,05g