`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)`

\(a) 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ b)n_{AlCl_3} = n_{Al} = \dfrac{8,1}{27} = 0,3(mol)\\ \Rightarrow m_{AlCl_3} = 0,3.133,5 = 40,05(gam)\\ c) n_{HCl} = 3n_{Al} = 0,9(mol)\\ \Rightarrow C_{M_{HCl}} = \dfrac{0,9}{0,5} = 1,8M\)

harigato pạn nhìu lắm lắm lắm....

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ a,Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{MgCl_2}=n_{H_2}=n_{Mg}=0,2\left(mol\right);n_{HCl}=0,2.2=0,4\left(mol\right)\\ b,C_{MddHCl}=\dfrac{0,4}{0,1}=4\left(M\right)\\ V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)

\(n_{AlCl_3}=\dfrac{26.7}{133.5}=0.2\left(mol\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(.........0.6......0.2.......0.3\)

\(V_{H_2}=0.3\cdot22.4=6.72\left(l\right)\)

\(C\%HCl=\dfrac{0.6\cdot36.5}{300}\cdot100\%=7.3\%\)

\(n_{Al} = \dfrac{10,8}{27} = 0,4(mol)\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ n_{AlCl_3} = n_{Al} = 0,4(mol)\\ m_{AlCl_3} = 0,4.133,5 = 53,4(gam)\\ n_{HCl} =3 n_{Al} = 1,2(mol)\\ C\%_{HCl}= \dfrac{1,2.36,5}{200}.100\% = 21,9\%\)

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

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

\(m_{HCl}=100.14,6\%=14,6\left(g\right)\Rightarrow n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,4}{2}\), ta được HCl dư.

Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{HCl\left(pư\right)}=2n_{Zn}=0,2\left(mol\right)\\n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{HCl\left(dư\right)}=0,4-0,2=0,2\left(mol\right)\)

Ta có: m _{dd sau pư} = 6,5 + 100 - 0,1.2 = 106,3 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,1.136}{106,3}.100\%\approx12,79\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,2.36,5}{106,3}.100\%\approx6,87\%\end{matrix}\right.\)

\(n_{Zn}=\dfrac{9.75}{65}=0.15\left(mol\right)\)

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

\(0.15.....0.3...................0.15\)

\(V_{H_2}=0.15\cdot22.4=3.36\left(l\right)\)

\(m_{HCl}=0.3\cdot36.5=10.95\left(g\right)\)

a) \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

b) \(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)

\(m_{dd}=m_{ct}+m_{dm}=7,2+150=157,2\left(g\right)\)

\(\Rightarrow C\%=\dfrac{7,2}{157,2}.100\%\approx4,6\%\)

c) Theo PTHH: \(n_{H_2}=n_{Mg}=0,3\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)