\(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\\ a,Fe+2HCl\rightarrow FeCl_2+H_2\\ b,n_{HCl}=0,4.2=0,8\left(mol\right)\\ m_{HCl}=0.8.36,5=29,2\left(g\right)\\ c,n_{H_2}=n_{FeCl_2}=n_{Fe}=0,4\left(mol\right)\\ m_{FeCl_2}=0,4.127=50,8\left(g\right)\\ d,V_{H_2\left(dktc\right)}=0,4.22,4=8,96\left(l\right)\)

a)

$n_{Fe} = \dfrac{14}{56} = 0,25(mol)$

$Fe + 2HCl \to FeCl_2 + H_2$

b) $n_{HCl} = 2n_{Fe} = 0,5(mol) \Rightarrow m_{HCl} = 0,5.36,5 = 18,25(gam)$

c) $n_{H_2} = n_{Fe} = 0,25(mol) \Rightarrow V_{H_2} = 0,25.22,4 = 5,6(lít)$

\(n_{Fe} =\dfrac{11,2}{56} = 0,2(mol)\\ \)

Fe + 2HCl → FeCl₂ + H₂

0,2.....0,4.........0,2........0,2..............(mol)

Vậy :

V = 0,2.22,4 = 4,48(lít)

\(m_{FeCl_2} = 0,2.127=25,4(gam)\)

\(m_{HCl} = 0,4.36,5 = 14,6(gam)\)

PTHH: Fe+2HCl → FeCl2+H2

a, nFe=m:M=11,2:56=0,2 mol

Theo PTHH, nFe=nH2=0,2 mol

VH2=n.22,4=0,2.22,4=4,48 lít

b, Theo PTHH, nFeCl2=nFe=0,2

mFeCl2=n.M=0,2.127=25,4 g 

c,

Theo PTHH, nHCl=2nFe=0,4 mol

mHCl=n.M=0,4.36,5=14,6 g

n_Fe=11,2/56=0,2(mol)

Fe + 2HCl ----->FeCl₂ + H₂

TPT:n_HCl=2.n_Fe=2.0,2=0,4(mol)

        m_HCl=0,4.36,5=14,6(g)

TPT:n_FeCl2=n_Fe=0,2(mol)

       m_FeCl2=0,2.127=25,4(g)

TPT:n_H2=n_Fe=0,2(mol)

       v_H2=0,2.22,4=4,48(lít)

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

\(Fe + 2HCl \to FeCl_2 + H_2\\ n_{Fe} = n_{FeCl_2} = n_{H_2} = \dfrac{8,96}{22,4} = 0,4(mol)\\ m_{Fe} = 0,4.56 = 22,4(gam)\\ m_{FeCl_2} = 0,4.127 = 50,8(gam)\)

\(n_{H_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)

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

\(n_{Fe}=n_{FeCl_2}=n_{H_2}=0.4\left(mol\right)\)

\(m_{Fe}=0.4\cdot56=22.4\left(g\right)\)

\(m_{FeCl_2}=0.4\cdot127=50.8\left(g\right)\)

a) PTHH : Fe + 2HCl → FeCl₂ + H₂

b) V_H2(đktc)=8,96(l) → n_H2= \(\dfrac{8,96}{22,4}\)=0,4(mol)

→ n_Fe= 0,4(mol)

m_Fe= 0,4 . 56=22,4(g)

c) Ta có :  n_H2= \(\dfrac{8,96}{22,4}\)=0,4(mol) → n_FeCl2= 0,4(mol)

m_FeCl2= 0,4 . 127=50,8(g)

`a)PTHH:`

`Fe + 2HCl -> FeCl_2 + H_2`

`0,4`                      `0,4`       `0,4`

`n_[H_2] = [ 8,96 ] / [ 22,4 ] = 0,4 (mol)`

`b) m_[Fe] = 0,4 . 56 = 22,4 (g)`

`c) m_[FeCl_2] = 0,4 . 127 = 50,8 (g)`

Fe+2HCl->FeCl2+H2

 0.4               0.4      0.4

nH2=V/22.4= 0.4 (mol)

mFe=n*M=0.4*56=22.4(g)

mFeCl2=n*M=0.4*(56+35.5*2)=50.8(g)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

Theo PT: \(\left\{{}\begin{matrix}n_{H_2}=n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\\n_{HCl}=2n_{Fe}=0,4\left(mol\right)\end{matrix}\right.\)

a, Ta có: \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

b, \(V_{ddHCl}=\dfrac{0,4}{1,5}\approx0,267\left(l\right)\)

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

Bạn tham khảo nhé!

\(n_{Cl}=n_{HCl}=2n_{H_2}=2\cdot\dfrac{8.96}{22.4}=0.8\left(mol\right)\)

\(m_{muối}=m_{kl}+m_{Cl}=11+0.8\cdot35.5=39.4\left(g\right)\)

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ n_{HCl}=\dfrac{18,25}{35}=0,5\left(mol\right)\\ a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{0,5}{2}>\dfrac{0,2}{1}\Rightarrow HCldư\\ n_{HCl\left(dư\right)}=0,5-0,2.2=0,1\left(mol\right)\\ m_{HCl}=0,1.36,5=3,65\left(g\right)\\ c.n_{ZnCl_2}=n_{Zn}=n_{H_2}=0,2\left(mol\right)\\ m_{ZnCl_2}=0,2.136=27,2\left(g\right)\\ V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

\(n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)

a) Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)

            1          2            1            1

           0,2       0,5         0,2         0,2

b) Lập tỉ số so sánh : \(\dfrac{0,2}{1}< \dfrac{0,5}{2}\)

                 ⇒ Zn phản ứng hết , HCl dư

                 ⇒ Tính toán dựa vào số mol của Zn

\(n_{HCl\left(dư\right)}=0,5-\left(0,2.2\right)=0,1\left(mol\right)\)

⇒ \(m_{HCl\left(dư\right)}=0,1.36,5=3,65\left(g\right)\)

c) \(n_{ZnCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{ZnCl2}=0,2.136=27,2\left(g\right)\)

\(n_{H2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)

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