200ml chứ

\(C_M=\dfrac{2}{\left(200.22,4\right):100}=0,046M\) 
 

`a) PTHH:`

`Mg + 2HCl -> MgCl_2 + H_2↑`

`0,5`        `1`             `0,5`        `0,5`        `(mol)`

`n_[Mg] = 12 / 24 = 0,5 (mol)`

   `=> V_[H_2] = 0,5 . 22,4 = 11,2 (l)`

   `=>m_[HCl] = 1 . 36,5 = 36,5(g)`

(a) Xét \(\dfrac{120}{250}.100=48\) => ddbh và có KCl không bị hoà tan hết

\(\left(b\right)m_{KCl\left(tan\right)}=\dfrac{250}{100}.42,6=106,5\left(g\right)\\ \rightarrow m_{KCl\left(ko.tan\right)}=120-106,5=13,5\left(g\right)\)

Khối lượng AgNO₃ có thể tan trong 250g nước ở 25^oC là:

\(\dfrac{250.222}{100}=555\left(g\right)\)

Em cảm ơn nhìu lắm:>

a.\(C\%_{KCl}=\dfrac{20}{600}.100=3,33\%\)

b.2,5kg = 2500g

\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{34,2}{2500}.100=1,368\%\)

`(a): C%_[KCl] = 20 / 600 . 100 ~~ 3,33 %`

________________________________________

`(b):` Đổi `2,5 kg = 2500 g`

`C%_[Al_2 (SO_4)_3] = [ 34,2 ] / 2500 . 100 = 1,368%`

a. Ta có: \(n_{CuSO_4}=\dfrac{400}{160}=2,5\left(mol\right)\)

\(\Rightarrow C_{M_{CuSO_4}}=\dfrac{2,5}{4}=0,625M\)

b. \(V_{dd_{BaCl_2}}=\dfrac{600}{1,2}=500\left(ml\right)=0,5\left(lít\right)\)

\(\Rightarrow C_{M_{BaCl_2}}=\dfrac{0,2}{0,5}=0,4M\)

\(\left(a\right)n_{CuSO_4}=\dfrac{400}{160}=2,5\left(mol\right)\\ C_{M\left(CuSO_4\right)}=\dfrac{2,5}{4}=0,525M\\ \left(b\right)V_{dd}=\dfrac{600}{1,2}=500\left(ml\right)\\ C_{M\left(BaCl_2\right)}=\dfrac{0,2}{0,5}=0,4M\)

Câu 3.

\(n_{K_2O}=\dfrac{2,35}{94}=0,025mol\)

\(K_2O+H_2O\rightarrow2KOH\)

0,025                   0,05   ( mol )

\(C_{M_{KOH}}=\dfrac{0,05}{0,75}=0,066M\)

Câu 4.

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

\(n_{HCl}=\dfrac{100.14,6\%}{36,5}=0,4mol\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

0,1   <  0,4                             ( mol )

0,1                          0,1               ( mol )

\(m_{ZnCl_2}=0,1.136=13,6g\)

cảm ơn ạ

a.b.

\(\left\{{}\begin{matrix}n_{Fe_2O_3}=40.80\%=32g\\m_{CuO}=40-32=8g\end{matrix}\right.\)

\(\left\{{}\begin{matrix}n_{Fe_2O_3}=\dfrac{32}{160}=0,2mol\\n_{CuO}=\dfrac{8}{80}=0,1mol\end{matrix}\right.\)

\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\)

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

\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)

0,2            0,6               0,4                  ( mol )

\(V_{H_2}=\left(0,1+0,6\right).22,4=15,68l\)

\(\left\{{}\begin{matrix}m_{Cu}=0,1.64=6,4g\\m_{Fe}=0,4.56=22,4g\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{6,4}{6,4+22,4}.100=22,22\%\\\%m_{Fe}=100\%-22,22\%=77,78\%\end{matrix}\right.\)

c.
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\) ( Cu không phản ứng với H₂SO₄ loãng )

0,4                                      0,4    ( mol )

\(V_{H_2}=0,4.22,4=8,96l\)

a, Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\)

\(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH:

Fe + 2HCl ---> FeCl₂ + H₂

a--->2a------------------>a

2Al + 6HCl ---> 2AlCl₃ + 3H₂

b---->3b-------------------->1,5b

=> \(\left\{{}\begin{matrix}56a+27b=16,6\\a+1,5b=0,5\end{matrix}\right.\Leftrightarrow a=b=0,2\left(mol\right)\)

=> \(\left\{{}\begin{matrix}m_{Fe}=0,2.56=11,2\left(g\right)\\m_{Al}=0,2.27=5,4\left(g\right)\end{matrix}\right.\)

b) \(C\%_{HCl}=\dfrac{\left(0,2.2+0,2.3\right).36,5}{300}.100\%=12,167\%\)

\(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\) 
gọi nFe : a , nAl: b (a,b>0)  => 56a + 27b = 16,6 (g) 
\(pthh:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\) 
           a                                     a
         \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\) 
           b                                     \(\dfrac{3b}{2}\)
          
=> \(a+\dfrac{3b}{2}=0,5\) 
ta có hệ pt 
\(\left\{{}\begin{matrix}56a+27b=16,6\\a+\dfrac{3b}{2}=0,5\end{matrix}\right.\) 
=> a= 0,2 , b = 0,2 
\(\left\{{}\begin{matrix}m_{Fe}=0,2.56=11,2\left(g\right)\\m_{Al}=16,6-11,2=5,4\left(g\right)\end{matrix}\right.\) 
\(pthh:Fe+2HCl\rightarrow FeCl_2+H_2\) 
           0,2     0,4 
        \(2Al+6HCl\rightarrow2AlCl_3+3H_2\) 
           0,2    0,6 
=> \(m_{HCl}=\left(0,4+0,6\right).36,5=36,5\left(g\right)\) 
=> \(C\%=\dfrac{36,5}{200}.100\%=18,25\%\)

a) \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

PTHH: 2Al + 6HCl ---> 2AlCl₃ + 3H₂

           0,1-->0,3------->0,1------>0,15

\(\rightarrow m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\)

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

c) PTHH: CuO + H₂ --t^o--> Cu + H₂O

             0,15<---0,15

=> m_CuO = 0,15.80 = 12 (g)

Cảm ơn bạn nhoa