a)

\(Mg + 2HCl \to MgCl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ Fe + 2HCl \to FeCl_2 + H_2\\ Zn + 2HCl \to ZnCl_2 + H_2\)

b)

Gọi : \(n_{H_2} = a(mol) \Rightarrow n_{HCl} = 2a\)

Bảo toàn khối lượng :

\(13,5 + 2a.36,5 = 66,75 + 2.a\\ \Rightarrow a = 0,75\\ \Rightarrow V = 0,75.22,4 = 16,8(lít)\)

a) Mg + 2 HCl -> MgCl2  + H2

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

Fe + 2 HCl ->  FeCl2 + H2

Zn + 2 HCl -> ZnCl2 + H2

b) mY-mX=mCl

<=> mCl= 66,75-13,5=53,25(g)

=>nCl=53,25/35,5=1,5(mol)

=> nH2= nCl/2= 1,5/2=0,75(mol)

=>V=V(H2,đktc)=0,75.22,4=16,8(l)

a) PTHH:

Mg + 2HCl ---> MgCl2 + H2 (1)

x x

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

\(\dfrac{4}{3}\)x 2x

Gọi nH2(1)= x => nH2 (2)= 2x

ta có mMg + mAl= 9

=> 24x + \(\dfrac{4}{3}\).27x =9

=> x + 0,15

mMg= 3,6 => mAl= 5,4

b)ΣnH2= 3x= 3.0,15= 0,45

=> nHCl= 0,45.2= 0,9

=> mHCl= 0,9.36,5= 32,85

a) PTHH:

(1) \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

\(n\) \(2n\) \(n\) \(n\) (mol)

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

\(\dfrac{4}{3}n\) \(4n\) \(\dfrac{4}{3}n\) \(2n\) (mol)

\(24n+\dfrac{4}{3}n.27=9\\ \Leftrightarrow60n=9\\ \Rightarrow n=\dfrac{9}{60}=0,15\left(mol\right)\)

\(\Rightarrow n_{Mg}=0,15\left(mol\right)\\ n_{Al}=\dfrac{4}{3}.0,15=0,2\left(mol\right)\)

b) \(m_{HCl}=36,5.\left(2n+4n\right)\)

\(=36,5\left(6.0,15\right)\\ =32,85\left(g\right)\)

a) Fe +2 HCl -> FeCl2 + H2

x____2x______x____x(mol)

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

y____3y______y________1,5y(mol)

b) nH2= 0,05(mol)

Ta có hpt:

\(\left\{{}\begin{matrix}56x+27y=1,66\\x+1,5y=0,05\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,02\\y=0,02\end{matrix}\right.\)

=> mFe=0,02.56= 1,12(g)

mAl=0,02.27=0,54(g)

Đặt x,y, z lần lượt là số mol của Na,Al,Mg trong m gam hỗn hợp A

m gam A + H2O dư

\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)

x--------------------x--------->0,5x

2Al + 2NaOH + 2H2O → 2NaAlO2 + 3H2

x<------x-------------------------------------->1,5x

=> \(0,5x+1,5x=\dfrac{2,24}{22,4}=0,1\left(mol\right)\) (1)

2m gam A + NaOH

\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)

2x------------------------------->x

2Al + 2NaOH + 2H2O → 2NaAlO2 + 3H2

2y---------------------------------------------->3y

=> \(x+3y=\dfrac{8,96}{22,4}=0,4\left(mol\right)\) (2) 

3m gam A + HCl

\(Na+HCl\rightarrow NaCl+\dfrac{1}{2}H_2\)

3x--------------------------->1,5x

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

3y----------------------------->4,5y

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

3z----------------------------->3z

=> \(1,5x+4,5y+3z=\dfrac{22,4}{22,4}=1\left(mol\right)\) (3)

Từ (1), (2), (3) =>\(\left\{{}\begin{matrix}x=0,05\\y=\dfrac{7}{60}\\z=\dfrac{2}{15}\end{matrix}\right.\)

=> \(m_{Na}=0,05.23=1,15\left(g\right)\)

\(m_{Al}=\dfrac{7}{60}.27=3,15\left(g\right)\)

\(m_{Mg}=\dfrac{2}{15}.24=3,2\left(g\right)\)

=> \(m=1,15+3,15+3,2=7,5\left(g\right)\)

=> \(\%m_{Na}=\dfrac{1,15}{7,5}.100=15,33\%\)

\(\%m_{Al}=\dfrac{3,15}{7,5}.100=42\%\)

\(\%m_{Mg}=\dfrac{3,2}{7,5}.100=42,67\%\)

\(2Na+2H2O\rightarrow2NaOH+H2\left(1\right)\)

\(2Al+2NaOH+2H2O\rightarrow2NaAlO2+3H2\left(2\right)\)

\(2Al+6HCl\rightarrow2AlCl3+3H2\left(3\right)\)

\(2Na+2HCl\rightarrow2NaCl+H2\left(4\right)\)

\(Mg+2HCl\rightarrow MgCl2+H2\left(5\right)\)

\(n_{H2\left(1\right)}=0,1\left(mol\right)\rightarrow n_{Na}=0,2\left(mol\right)\rightarrow m_{Na}=4,6\left(g\right)\)

\(n_{H2\left(2\right)}=0,4\left(mol\right)\Rightarrow n_{Al}=\dfrac{4}{15}\left(mol\right)\Rightarrow m_{Al}=7,2\left(g\right)\)

\(\Rightarrow n_{H2\left(3\right)}=\dfrac{3}{2}n_{Al}=0,4\left(mol\right)\)

\(n_{H2\left(4\right)}=\dfrac{1}{2}n_{Na}=0,1\left(mol\right)\)

\(\Rightarrow n_{H2\left(5\right)}=1-0,4-0,1=0,5\left(mol\right)\)

\(\Rightarrow n_{Mg}=0,5\left(mol\right)\Rightarrow m_{Mg}=12\left(g\right)\)

\(\Rightarrow m=12+4,6+7,2=23,8\left(g\right)\)

\(\%m_{Na}=\dfrac{4,6}{23,8}.100\%=19,33\%\)

\(\%m_{Al}=\dfrac{7,2}{23,8}.100\%=30,25\%\)

\(\%m_{Mg}=100-19,33-30,25=50,42\%\)

Chúc bạn học tốt

a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)

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

b, 24n_Mg + 27n_Al = 6,12 (1)

Theo PT: \(n_{H_2}=n_{Mg}+\dfrac{3}{2}n_{Al}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,12\left(mol\right)\\n_{Al}=0,12\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,12.24}{6,12}.100\%\approx47,06\%\\\%m_{Al}\approx52,94\%\end{matrix}\right.\)

\(a.2Al+6HCl->2AlCl_3+3H_2\\ Mg+2HCl->MgCl_2+H_2\\ b.n_{Al}=a,n_{Mg}=b\\ 27a+24b=7,5\left(I\right)\\ 1,5a+b=\dfrac{7,84}{22,4}=0,35\left(II\right)\\ a=0,1;b=0,2\\ \%m_{Al}=\dfrac{27\cdot0,1}{7,5}\cdot100\%=36\%\\ \%m_{Mg}=64\%\\ c.m_{HCl}=36,5\left(0,1\cdot3+0,2\cdot2\right)=18,25g\\ d.m_{ddsau}=7,5+\dfrac{18,25}{14,6:100}-0,35\cdot2=131,8g\\ C\%\left(AlCl_3\right)=\dfrac{133,5\cdot0,1}{131,8}\cdot100\%=10,1\%\\ C\%\left(MgCl_2\right)=\dfrac{95\cdot0,2}{131,8}\cdot100\%=14,4\%\)

a) PTHH : Mg + 2HCl \(\rightarrow\) MgCl₂ + H₂\(\uparrow\) (1)

MgO + 2HCl \(\rightarrow\) MgCl₂ + H₂O(2)

b) Theo đề cho : n_H2=4.48/22.4=0.2(mol)

Theo PT(1): n_Mg=n_H2=0.2(mol)

Do đó m_Mg(A)=0.2 \(\times\)24 =4.8(g)

m_MgO(A) = 8.8-4,8=4(g)

c) Ta có : n_MgO = 4/40 =0.1(mol)

Theo các PT(1)(2):

\(\Sigma\)n_HCl(p/ư) = 2 \(\times\)(0.2 +0.1) =0.6(mol)

\(\Rightarrow\)V_{HCl ^{= \(\dfrac{0.6}{2}\)=0.3(l}}ít)