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

\(n_{Fe}=\dfrac{m}{M}=\dfrac{28}{56}=0,5\left(mol\right)\\ PTHH:Fe+2HCl->FeCl_2+H_2\)

ti le        1      :    2       :      1      :    1

n(mol)     0,5-->1--------->0,5------>0,5

\(m_{FeCl_2}=n\cdot M=0,5\cdot\left(56+35,5\cdot2\right)=63,5\left(g\right)\\ V_{H_2\left(dktc\right)}=n\cdot22,4=0,5\cdot22,4=11,2\left(l\right)\)

\(a)n_{Fe}=\dfrac{5,6}{56}=0,1mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\)

0,1       0,2            0,1          0,1

\(m_{FeCl_2}=0,1.127=12,7g\\ b)V_{H_2}=0,1.22,4=2,24l\\ c)C_{M_{HCl}}=\dfrac{0,2}{0,2}=1M\)

Pt: Fe + 2HCl => FeCl₂ + H₂

nFe = \(\dfrac{2,8}{56}=0,05mol\)

nHCl = \(\dfrac{14,6}{36,5}=0,4mol\)

nFe : nHCl = \(\dfrac{0,05}{1}:\dfrac{0,4}{2}=0,05:0,2=1:4\)

=> HCl dư

a) nFeCl₂ = nFe = 0,05 mol => mFeCl₂ = 6,35g

b) nH₂ = nFe = 0,05 mol => VH₂ = 0,05.22,4 = 1,12 lít

$a) Fe + 2HCl \to FeCl_2 + H_2$

$b) n_{Fe} = \dfrac{5,6}{56} = 0,1(mol)$
Theo PTHH : $n_{HCl} = 2n_{Fe} = 0,1.2 = 0,2(mol)$
$m_{HCl} = 0,2.36,5 = 7,3(gam)$
$c) n_{H_2} = n_{Fe} = 0,1(mol)$
$V_{H_2} = 0,1.22,4 = 2,24(lít)$

a: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

b: \(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)=n_{FeCl_2}\)

\(\Leftrightarrow n_{HCl}=2\cdot0.1=0.2\left(mol\right)\)

\(m=0.2\cdot36.5=7.3\left(g\right)\)

c: \(V_{H_2}=0.1\cdot22.4=2.24\left(lít\right)\)

\(a. Fe+2HCl->FeCl_2+H_2\)

c, \(n_{Fe}=\dfrac{5,6}{56}=0,1\)

\(-> n_{H_2}=n_{FeCl_2}=0,1\)

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

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

Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

____0,1______0,2_____0,1____0,1 (mol)

a, \(C_{M_{HCl}}=\dfrac{0,2}{0,15}=\dfrac{4}{3}\left(M\right)\)

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

b, \(FeCl_2+2NaOH\rightarrow2NaCl+Fe\left(OH\right)_2\)

Theo PT: \(n_{NaOH}=2n_{FeCl_2}=0,2\left(mol\right)\)

\(\Rightarrow V_{NaOH}=\dfrac{0,2}{2}=0,1\left(l\right)\)

a. PTHH : Fe + 2HCl -> FeCl2 + H2
b. Số mol Fe = 11,2/56 = 0,2 mol. Số mol HCl = 0,4 mol nên m(HCl) = 36,5.0,4 = 14,6 g.
Số mol FeCl2 = số mol H2 = số mol Fe = 0,2 mol.
m(FeCl2) = 127.0,2 = 25,4 g
=> V(H2) = 0,2.22,4 = 4,48 lít.

\(a.Fe+2HCl\rightarrow FeCl_2+H_2\\b.n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ n_{H_2}=n_{Fe}=0,1\left(mol\right)\\ \Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\\ c.n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\\ m_{FeCl_2}=0,1.127=12,7\left(g\right) \)