n H2 = 13,44/22,4 = 0,6(mol)

$2Al + 6HCl \to 2AlCl_3 + 3H_2$

n Al = 2/3 n H2 = 0,4(mol)

%m Al = 0,4.27/21  .100% = 51,43%
%m Al2O3 = 100% -51,43% = 48,57%

b)

=> n Al2O3 = (21 - 0,4.27)/102 = 0,1(mol)

$Al_2O_3 + 6HCl \to 2AlCl_3 + 3H_2O$

n HCl = 3n Al + 6n HCl = 0,4.3 + 0,1.6 = 1,8(mol)

=> m dd HCl = 1,8.36,5/20% = 328,5(gam)

=> V dd HCl = m/D = 328,5/1,18 = 278,39(ml)

Gọi n Fe = a (mol )

       n Mg = b (mol )  (a,b > 0)

--> 56a+24b = 13,2 

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

a            2a           a           a

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

b             2b             b            b

----> a+b=0,35 

Ta có hệ Pt :

\(\left\{{}\begin{matrix}56a+24b=13,2\\a+b=0,35\end{matrix}\right.\)

Giải hệ PT , ta có : 

a= 0,15 

b = 0,2 (mol )

\(V_{HClđủ}=\left(0,15.2+0,2.2\right):0,5=1,4\left(l\right)\)

\(a,m_{Fe}=0,15.56=8,4\left(g\right)\)

\(m_{Mg}=0,2.24=4,8\left(g\right)\)

\(\%m_{Fe}=\dfrac{8,4}{13,2}.100\%\approx63,64\%\)

\(\%m_{Mg}=\dfrac{4,8}{13,2}.100\%\approx36,36\%\)

\(b,m_{FeCl_2}=0,15.127=19,05\left(g\right)\)

\(m_{MgCl_2}=0,2.95=19\left(g\right)\)

\(c,HCl+NaOH\rightarrow NaCl+H_2O\)

    0,2        0,2 

\(m_{NaOH}=\dfrac{100.8}{100}=8\left(g\right)\)

\(n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)

\(V_{HCldư}=\dfrac{n}{C_M}=\dfrac{0,2}{0,5}=0,4\left(l\right)\)

\(V_{HCl}=V_{HClđủ}+V_{HCldư}=1,4+0,4=1,8\left(l\right)\)

\(2\left[H\right]+\left[O\right]->H_2O\\ n_{Cl}=n_H=2n_O=\dfrac{44,6-28,6}{16}.2=2mol\\ m_{muoi}=28,6+35,5.2=99,6g\)

a) Do sản phẩm thu được sau khi nung khi hòa tan vào dd HCl thu được hỗn hợp khí => Sản phẩm chứa Fe dư

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\); \(n_S=\dfrac{1,6}{32}=0,05\left(mol\right)\)

PTHH: Fe + S --t^o--> FeS 

          0,05<-0,05-->0,05

            Fe + 2HCl --> FeCl₂ + H₂

          0,05-->0,1---------->0,05

             FeS + 2HCl --> FeCl₂ + H₂S

            0,05-->0,1------------->0,05

=> \(\%V_{H_2}=\%V_{H_2S}=\dfrac{0,05}{0,05+0,05}.100\%=50\%\)

b)

n_NaOH = 0,2.1 = 0,2 (mol)

PTHH: NaOH + HCl --> NaCl + H₂O

                0,2-->0,2

=> n_HCl = 0,1 + 0,1 + 0,2 = 0,4 (mol)

=> \(C_{M\left(dd.HCl\right)}=\dfrac{0,4}{0,5}=0,8M\)

a, PT: \(Fe+S\underrightarrow{t^o}FeS\)

Ta có: \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)

\(n_S=\dfrac{3,2}{32}=0,1\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,3}{1}>\dfrac{0,1}{1}\), ta được Fe dư.

Chất rắn A gồm Fe dư và FeS.

Theo PT: \(n_{Fe\left(pư\right)}=n_{FeS}=n_S=0,1\left(mol\right)\)

\(\Rightarrow n_{Fe\left(dư\right)}=0,2\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(FeS+2HCl\rightarrow FeCl_2+H_2S\)

Theo PT: \(\left\{{}\begin{matrix}n_{H_2}=n_{Fe\left(dư\right)}=0,2\left(mol\right)\\n_{H_2S}=n_{FeS}=0,1\left(mol\right)\end{matrix}\right.\)

Ở cùng điều kiện nhiệt độ và áp suất, %V cũng là % số mol.

\(\Rightarrow\left\{{}\begin{matrix}\%V_{H_2}=\dfrac{0,2}{0,2+0,1}.100\%\approx66,67\%\\\%V_{H_2S}\approx33,33\%\end{matrix}\right.\)

b, Ta có: \(\Sigma n_{HCl\left(dadung\right)}=2n_{Fe}+2n_{FeS}=0,6\left(mol\right)\) (1)

PT: \(NaOH+HCl\rightarrow NaCl+H_2O\)

Ta có: \(n_{NaOH}=0,1.2=0,2\left(mol\right)\)

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

Từ (1) và (2) \(\Rightarrow\Sigma n_{HCl}=0,8\left(mol\right)\)

\(\Rightarrow C_{M_{ddHCl}}=\dfrac{0,8}{0,5}=1,6\left(M\right)\)

Bạn tham khảo nhé!

a) \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

Gọi số mol Zn, Al là a, b

=> 65a + 27b = 18,4 (1)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

           a----->2a------------->a

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

            b---->3b------------->1,5b

=> a + 1,5b = 0,5 (2)

(1)(2) => a = 0,2 ; b = 0,2

=> \(\left\{{}\begin{matrix}m_{Zn}=0,2.65=13\left(g\right)\\m_{Al}=0,2.27=5,4\left(g\right)\end{matrix}\right.\)

b) n_HCl(pư) = 2a + 3b = 1 (mol)

n_HCl(dư) = 0,6.2 - 1 = 0,2 (mol)

PTHH: KOH + HCl --> KCl + H₂O

           0,2<----0,2

=> \(V=\dfrac{0,2}{1}=0,2\left(l\right)\)

\(n_{HCl}=0,6.2=1,2\left(mol\right)\\ n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\ a,n_{HCl\left(dư\right)}=1,2-2.n_{H_2}=1,2-2.0,5=0,2\left(mol\right)\\PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Đặt:n_{Zn}=a\left(mol\right);n_{Al}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}65a+27b=18,4\\a+1,5b=0,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,2\end{matrix}\right.\\ \Rightarrow m_{Zn}=0,2.65=13\left(g\right);m_{Al}=0,2.27=5,4\left(g\right)\\ b,KOH+HCl_{dư}\rightarrow KCl+H_2O\\ n_{KOH}=n_{HCl\left(dư\right)}=0,2\left(mol\right)\\ \Rightarrow V=V_{ddKOH}=\dfrac{0,2}{1}=0,2\left(l\right)\)

a)\(\left\{{}\begin{matrix}n_{Zn}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\)

\(n_{HCl}=V\cdot C_M=2\cdot0,6=1,2mol\)

\(n_{\uparrow}=\dfrac{11,2}{22,4}=0,5mol\)

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

 x         2x           x            x

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

 y         3y           y               1,5y

Ta có hệ: \(\left\{{}\begin{matrix}65x+27y=18,4\\x+1,5y=0,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,2\end{matrix}\right.\)

\(m_{Zn}=0,2\cdot65=13g\)

\(m_{Al}=0,2\cdot27=5,4g\)

b)Theo pt: \(\Sigma n_{HCl}=2x+3y=2\cdot0,2+3\cdot0,2=1mol\)

Để trung hòa: \(n_{OH^-}=n_{H^+}=1mol\)

\(\Rightarrow V=\dfrac{n_{OH^-}}{C_M}=\dfrac{1}{1}=1l\)