`Fe + 2HCl -> FeCl_2 + H_2 \uparrow`

 `0,1`      `0,2`          `0,1`       `0,1`         `(mol)`

`n_[Fe]=[5,6]/56=0,1(mol)`

`a)V_[H_2]=0,1.22,4=2,24(l)`

`b)C_[M_[HCl]]=[0,2]/[0,1]=2(M)`

a)

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

b),c)

Theo PTHH :

\(n_{ZnCl_2} = n_{H_2} = n_{Zn} = \dfrac{13}{65} = 0,2(mol)\)

Vậy : 

\(m_{ZnCl_2} = 0,2.136 = 27,2(gam)\\ V_{H_2} =0,2.22,4 = 4,48(lít)\)

a) PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

b+c)

Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)=n_{ZnCl_2}=n_{H_2}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{ZnCl_2}=0,2\cdot136=27,2\left(g\right)\\V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\end{matrix}\right.\)

a. Zn + 2HCl → ZnCl₂ + H₂

b. n_Zn = n\(_{ZnCl_2}\) =\(\dfrac{13}{65}=0,2\left(mol\right)\) => m\(_{ZnCl_2}\)= 0,2.136 = 27,2(g)

c. n\(_{H_2}\)= n_Zn = 0,2 (mol) => V\(_{H_2}\)=0,2.22,4 = 4,48 (lít)

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

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

 1           1              1           1    (mol)

0,2       0,2             0,2        0,2    (mol)

\(VH_2=0,2.22,4=4,48\left(l\right)\)

\(Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\)

 1             3         2             3    (mol)

              0,2         2/15            (mol)

\(mFe=\dfrac{2}{15}.56=7,47\left(g\right)\)

Oke

a) $Zn + H_2SO_4 \to ZnSO_4 + H_2$

b) Theo PTHH :

$n_{H_2} = n_{ZnSO_4} = n_{Zn} = \dfrac{32,5}{65} = 0,5(mol)$

$V_{H_2} = 0,5.22,4 = 11,2(lít)$

$m_{ZnSO_4} = 0,5.161 = 80,5(gam)$

c) 

$n_{H_2SO_4} = n_{Zn} = 0,5(mol)$

$C\%_{H_2SO_4} = \dfrac{0,5.98}{168,5}.100\% = 29,08\%$

\(n_{Zn}=\dfrac{26}{65}=0.4\left(mol\right)\)

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

\(0.4.......0.8....................0.4\)

\(m_{HCl}=0.8\cdot36.5=29.2\left(g\right)\)

\(V_{H_2}=0.4\cdot22.4=8.96\left(l\right)\)

\(a.PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\)

\(b.n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

Từ PTHH trên ta có:

1 mol Mg tác dụng với 2 mol HCl sinh ra 1 mol MgCl₂ và 1 mol H₂ 

=> 0,1 mol Mg tác dụng với 0,2 mol HCl sinh ra 0,1 mol MgCl₂ và 0,1 mol H₂ 

\(\Rightarrow m_{HCl}=36,5.0,2=7,3\left(g\right)\)

\(c.\Rightarrow m_{H_2}=0,1.2=0,2\left(g\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\)

\(d.C_1:\\ Áp.dụng.ĐLBTKL,ta.có:\\ m_{Mg}+m_{HCl}=m_{MgCl_2}+m_{H_2}\\ \Rightarrow m_{MgCl_2}=m_{Mg}+m_{HCl}-m_{H_2}=2,4+7,3-0,2=9,5\left(g\right)\)

\(C_2:Từ.PTHH.trên.ta.có:\\ n_{MgCl_2}=0,1\left(mol\right)\\ \Rightarrow m_{MgCl_2}=0,1.95=9,5\left(g\right)\)

a) 2Al + 6HCl -> 2AlCl3 + 3H2

b)nAl = 2.7/27=0.1mol

2Al + 6HCl -> 2AlCl3 + 3H2

(mol) 0.1 0.3 0.1 0.15

VH2 = 0.15*22.4=3.36l

mAlCl3= 0.1*133.5=13.35g

=>C%=\(\dfrac{13,35}{500+27}100=2,533\%\)

a) Pt: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

b) nFe = \(\dfrac{11,2}{56}=0,2mol\)

Theo pt: nH₂ = nFe = 0,2 mol

=> VH₂ = 0,2.22,4 = 4,48lit

c) Theo pt: nHCl = 2nFe = 0,4 mol

=> mHCl = 0,4.36,5 = 14,6 g

=> C% = \(\dfrac{14,6}{73}.100\%=20\%\)