`a)PTHH:`

`Fe + 2HCl -> FeCl_2 + H_2`

`0,3`    `0,6`          `0,3`     `0,3`       `(mol)`

`n_[Fe]=[22,4]/56=0,4(mol)`

`n_[HCl]=0,3.2=0,6(mol)`

Ta có:`[0,4]/1 > [0,6]/2`

   `=>Fe` dư

`b)m_[FeCl_2]=0,3.127=38,1(g)`

`c)m_[Fe(dư)]=(0,4-0,3).56=5,6(g)`

\(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\\ n_{HCl}=0,3.2=0,6\left(mol\right)\\ pthh:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\\ LTL:\dfrac{0,4}{1}>\dfrac{0,6}{2}\) 
=> Fe dư 
\(n_{FeCl_2}=\dfrac{1}{2}n_{HCl}=0,3\left(mol\right)\\ m_{FeCl_2}=0,3.127=38,1\left(g\right)\\ n_{Fe\left(p\text{ư}\right)}=\dfrac{1}{2}n_{HCl}=0,3\left(mol\right)\\ m_{Fe\left(d\right)}=\left(0,4-0,3\right).56=5,6\left(g\right)\)

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

\(n_{HCl}=0,3.2=0,6\left(mol\right)\)

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

Xét: \(\dfrac{0,4}{1}>\dfrac{0,6}{2}\)                            ( mol )

         0,3      0,6          0,3               ( mol )

\(m_{FeCl_2}=0,3.127=38,1\left(g\right)\)

\(m_{Fe\left(dư\right)}=\left(0,4-0,3\right).56=5,6\left(g\right)\)

a) nFe=0,1(mol); nHCl=0,4(mol)

PTHH: Fe + 2 HCl -> FeCl2 + H2

Ta có: 0,1/1 < 0,4/2

=> Fe hết, HCl dư, tish theo nFe.

b) nH2=nFeCl2=Fe=0,1(mol)

=> V(H2,đktc)=0,1.22,4=2,24(l)

c) mFeCl2=127.0,1=12,7(g)

a) nFe=0,1(mol); nHCl=0,4(mol) PTHH: Fe + 2 HCl -> FeCl2 + H2 Ta có: 0,1/1 < 0,4/2 => Fe hết, HCl dư, tish theo nFe. b) nH2=nFeCl2=Fe=0,1(mol) => V(H2,đktc)=0,1.22,4=2,24(l) c) mFeCl2=127.0,1=12,7(g)

\(a,\text{Sơ đồ p/ứ: }Fe+HCl\to FeCl_2+H_2\\ b,PTHH:Fe+2HCl\to FeCl_2+H_2\\ c,\text{Bảo toàn KL: }m_{Fe}+m_{HCl}=m_{FeCl_2}+m_{H_2}\\ \Rightarrow m_{HCl}+56=150+8=158\\ \Rightarrow m_{HCl}=102(g)\)

Fe + 2HCl –> FeCl2 + H2

a) Số mol Fe: 2,8/56=0,05 (mol)

Theo pthh số mol H2= 0,05 mol

Thể tích H2=0,05 x 22,4 = 1.12 (lít)

b) Theo pthh số mol HCl= 0,1 mol

Khối lượng HCl cần dùng: 0,1 x 36,5 = 3,65 (g)

trên là bài 1 đây là bài 2 nha

Số mol Al là: 

Bài 1:

a, Fe + 2HCl \(\rightarrow\) FeCl₂ + H₂\(\uparrow\) ( 1:2:1:1 )

b, 

 \(n_{Fe}=\dfrac{2,8}{56}=0,05mol\\ \Rightarrow n_{H_2}=0,05.1=0,05mol\\ \Rightarrow V_{H_2}=0,05.22,4=1,12l\)

c, \(n_{HCl}=0,05.2=0,1mol\\ \Rightarrow m_{HCl}=0,1.36,5=3,65g\)

Bài 2:

a, 2Al + 6HCl \(\rightarrow\) 2AlCl₃ + 3H₂ ( 2:6:2:3 )

b, \(n_{Al}=\dfrac{5,4}{27}=0,2mol\)

\(\Rightarrow n_{H_2}=0,2.3=0,6mol\)

\(V_{H_2}=0,6.22,4=13,44l\)

c, \(n_{HCl}=0,2.6=1,2mol\\ \Rightarrow m_{HCl}=1,2.36,5=43,8g\)

Fe + 2HCl --> FeCl₂ + H₂

a. sắt + axit clohydric -> sắt(II) clorua + hidro

b. \(Fe+2HCl->FeCl_2+H_2\)

\(N_{HCl}:N_{FeCl_2}:N_{H_2}=2:1:1\\ c.BTKL:m_{ddHCl}=205,4+0,2-5,6=200\left(g\right)\)

a

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

0,1                                 0,1

b

Số nguyên tử Fe: số phân tử `HCl`: số phân tử `FeCl_2`: số phân tử `H_2` là `1:2:1:1`

c

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ n_{H_2}=\dfrac{0,2}{2}=0,1\left(mol\right)\)

Do \(n_{Fe}=n_{H_2}\) nên `Fe` không dư.

\(m_{dd.FeCl_2}=m_{Fe}+m_{dd.HCl}-m_{H_2}\\ \Leftrightarrow205,4=5,6+m_{dd.HCl}-0,2\)

\(\Rightarrow m_{dd.HCl}=200\left(g\right)\)

nHCl=2.0,1=0,2mol

PTHH: Fe+2HCl=>FeCl2+H2

           0,1<-0,2--->0,1---->0,1

mFe phản ứng=0,1.56=5,6g

VH2 thu đc=0,1.22,4=2,24 lít

a. \(Fe+2HCl\rightarrow FeCl_2+H_2\)

b. \(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{28}{56}=0,5\left(mol\right)\)

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

0,5-------1---------0,5-----0,5

Theo PTHH: \(\Rightarrow n_{H_2}=n_{Fe}=0,5\left(mol\right)\)

\(V_{H_2}=n_{H_2}.22,4=0,5.22,4=11,2\left(l\right)\)

c. \(H_2+CuO\rightarrow Cu+H_2O\)

  0,5-------0,5-----0,5----0,5

\(\Rightarrow m_{Cu}=n_{Cu}.M_{Cu}=0,5.64=32\left(g\right)\)

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

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

0,1  < 0,4                              ( mol )

0,1                                    0,1   ( mol )

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

\(n_{ZnCl_2}=\dfrac{0,1.1}{1}=0,1mol\)