gọi x,y lần lượt là số mol của Al và Fe

PTHH 2Al + 6HCl ---> 2AlCl₃ + 3H₂

x 3x x 1,5x ( mol)

Fe + 2HCl ----> FeCl₂ + H₂

y 2y y y (mol)

ta có hệ PT : \(\left\{{}\begin{matrix}27x+56y=5,5\\1,5x+y=\dfrac{4,48}{22,4}=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,05\left(mol\right)\end{matrix}\right.\)

=> m_Al = n . M = 0,1 . 27 = 2,7 (g)

=> m_Fe = n . M = 0,05 . 56 =2,8

=> \(\%m_{Al}=\dfrac{2,7}{5,5}\cdot100=49,09\%\)

\(\rightarrow\%m_{Fe}=100-49,09=50,91\%\)

b) ta có : \(\sum n_{HCl}=3x+2y=0,4\left(mol\right)\Rightarrow m_{HCl}=0,4\cdot36,5=14,6\left(g\right)\)

\(\Rightarrow mdd_{HCl}=\dfrac{14,6\cdot100}{14,6}=100\left(g\right)\)

mdd_{sau phản ứng} = m_Al+Fe + mdd_HCl - m_H2

= 2,8 +2,7 + 100 - 0,2 . 2

= 105,1 (g)

\(m_{AlCl_3}=n\cdot M=0,1\cdot133,5=13,35\Rightarrow C\%=\dfrac{13,15}{105,1}\cdot100\%=12,7\%\)

\(m_{FeCl_2}=n.M=0,05\cdot127=6,35\Rightarrow C\%=\dfrac{6,35}{105,1}\cdot100=6,04\%\)

n H2 = 0,2 mol.
2Al + 6HCl -> 2AlCl3 + 3H2
a -> ....3a...........a........1,5a (mol)

Fe + 2HCl -> FeCl2 + H2
b -> ...2b..........b.........b (mol)

Theo đề bài, ta có hpt:
27a + 56b = 5,5
1,5a + b = 0,2
Giải hệ, được: a = 0,1; b = 0,05.

a/ m Al = 27a = 2,7g.
=> %Al = 49,09 %.
=> % Fe = 50,91 %. .

b/ m AlCl3 = 13,35g.
m FeCl2 = 6,35g.
Áp dụng đ. luật bảo toàn khối lượng:
mdd sau = mdd trước - m khí
.............= mdd HCl - m H2
.............= 100 - 2.0,2
.............= 99,6g
=> C% AlCl3 = 13,4 %.
.....C% FeCl2 = 6,38 %.

PTHH :

\(Al+3HCl\rightarrow AlCl_3+1,5H_2\uparrow\)

x 3x

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

x 2x

Gọi số mol của Al là x => \(m_{Al}=27x\)

Gọi số mol của Fe là y => \(m_{Fe}=56y\)

Ta thấy cứ 1 mol HCl lại tạo ra 0,5 H₂

Mà : \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

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

Ta có HPT : \(\left\{{}\begin{matrix}27x+56y=5,5\\3x+2y=0,4\end{matrix}\right.\)

Giải HPT ta được : \(x=0,1\left(mol\right),y=0,2\left(mol\right)\)

Ta có : \(m_{Al}=2,7\left(g\right)\)

\(m_{Fe}=11,2\left(g\right)\)

a ) \(\%Al=\dfrac{2,7}{11,2+2,7}.100\%\approx19,4\%\)

\(\Rightarrow\%Fe=100\%-19,4\%=80,6\%.\)

b ) Ta có : \(m_{HCl}=14,6\left(g\right)\)

Ta có : \(m_{dd\left(HCl\right)}=\dfrac{m_{HCl}}{C\%}=\dfrac{14,6}{\dfrac{14,6}{100}}=100\left(ml\right)\)

m\(m_{H_2}=0,4.2=0,8\)

\(m_{dd\left(sau\right)}=100+11,2+2,7-0,8=113,1\left(ml\right)\)

\(m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\)

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

\(C\%=\dfrac{25,4+13,35}{113,1}.100\%\approx34,26\%.\)

a) Gọi số mol Mg, Al, Fe trong m gam hỗn hợp là a, b, c (mol)

\(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)

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

_______a--------------------->a------->a_______(mol)

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

_b-------------------->b------->1,5b___________(mol)

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

_c------------------>c------->c_______________(mol)

=> \(\left\{{}\begin{matrix}a+1,5b+c=0,35\left(1\right)\\95a+133,5b+127c=35,55\left(2\right)\end{matrix}\right.\)

Mặt khác:

PTHH: \(Mg+Cl_2\underrightarrow{t^o}MgCl_2\)

_______a--------------->a_________(mol)

\(2Al+3Cl_2\underrightarrow{t^o}2AlCl_3\)

_b----------------->b______________(mol)

\(2Fe+3Cl_2\underrightarrow{t^o}2FeCl_3\)

_c------------------>c______________(mol)

=> 95a + 133,5b + 162,5 = 39,1 (3)

(1)(2)(3) => \(\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,1\left(mol\right)\\c=0,1\left(mol\right)\end{matrix}\right.\) 

=> m = 24.0,1 + 27.01 + 56.0,1 = 10,7(g)

b) \(\left\{{}\begin{matrix}m_{Mg}=24.0,1=2,4\left(g\right)\\m_{Al}=27.0,1=2,7\left(g\right)\\m_{Fe}=56.0,1=5,6\left(g\right)\end{matrix}\right.\)

\(\text{Đặt }n_{Al}=x(mol);n_{Fe}=y(mol)\\ \Rightarrow 27x+56y=13,9(1)\\ n_{H_2}=\dfrac{7,84}{22,4}=0,35(mol)\\ a,PTHH:2Al+6HCl\to 2AlCl_3+3H_2(1)\\ Fe+2HCl\to FeCl_2+H_2(2)\\ b,\text{Từ 2 PT: }1,5x+y=0,35(2)\\ (1)(2)\Rightarrow x=0,1(mol);y=0,2(mol)\\ \Rightarrow m_{Al}=0,1.27=2,7(g)\\ m_{Fe}=0,2.56=11,2(g)\)

\(c,n_{HCl(1)}=3n_{Al}=0,3(mol);n_{AlCl_3}=0,1(mol);n_{H_2(1)}=0,15(mol)\\ \Rightarrow m_{dd_{HCl(1)}}=\dfrac{0,3.36,5}{14,6\%}=75(g)\\ \Rightarrow C\%_{AlCl_3}=\dfrac{0,1.133,5}{2,7+75-0,15.2}.100\%=17,25\%\)

\(n_{HCl(2)}=2n_{Fe}=0,4(mol);n_{FeCl_2}=n_{H_2(2)}=n_{Fe}=0,2(mol)\\ \Rightarrow m{dd_{HCl(2)}}=\dfrac{0,4.36,5}{14,6\%}=100(g)\\ \Rightarrow C\%_{FeCl_2}=\dfrac{0,2.127}{11,2+100-0,2.2}.100\%=22,92\%\)

a) 2Al + 6HCl --> 2AlCl₃ + 3H₂

Fe + 2HCl --> FeCl₂ + H₂

b) Gọi số mol Al, Fe lần lượt là a,b 

=> 27a + 56b = 13,9

\(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)

2Al + 6HCl --> 2AlCl₃ + 3H₂

a----->3a--------->a------->1,5a______(mol)

Fe + 2HCl --> FeCl₂ + H₂

b------>2b-------->b----->b__________(mol)

=> 1,5a + b = 0,35

=> \(\left\{{}\begin{matrix}a=0,1=>m_{Al}=0,1.27=2,7\left(g\right)\\b=0,2=>m_{Fe}=0,2.56=11,2\left(g\right)\end{matrix}\right.\)

c) n_HCl = 3a + 2b = 0,7 (mol)

=> m_HCl = 0,7.36,5 = 25,55(g)

=> \(m_{ddHCl}=\dfrac{25,55.100}{14,6}=175\left(g\right)\)

\(m_{dd\left(saupu\right)}=13,9+175-2.0,35=188,2\left(g\right)\)

\(\left\{{}\begin{matrix}m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\\m_{FeCl_2}=0,2.127=25,4\left(g\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}C\%\left(AlCl_3\right)=\dfrac{13,35}{188,2}.100\%=7,1\%\\C\%\left(FeCl_2\right)=\dfrac{25,4}{188,2}.100\%=13,5\%\end{matrix}\right.\)

a) \(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

PTHH: 2Al + 6HCl → 2AlCl₃ + 3H₂

Mol:       x                                  1,5x

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:       y                                 y

Ta có: \(\left\{{}\begin{matrix}27x+65y=24,9\\1,5x+y=0,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}27x+65.\left(0,6-1,5x\right)=24,9\\y=0,6-1,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,3\end{matrix}\right.\)

b, \(m_{Al}=0,2.27=5,4\left(g\right);m_{Zn}=24,9-5,4=19,5\left(g\right)\)

c) \(\%m_{Al}=\dfrac{5,4.100\%}{24,9}=21,69\%;\%m_{Zn}=100\%-21,69\%=78,31\%\)

d) 

PTHH: 2Al + 6HCl → 2AlCl₃ + 3H₂

Mol:      0,2     0,6          0,2       0,3

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:      0,3     0,6        0,3       0,3

\(m_{ddHCl}=\dfrac{\left(0,6+0,6\right).36,5.100}{14}=312,857\left(g\right)\)

e) m_{dd sau pứ} = 24,9 + 312,857 - (0,3+0,3).2 = 336,557 (g)

\(C\%_{ddAlCl_3}=\dfrac{0,2.133,5.100\%}{336,557}=7,93\%\)

\(C\%_{ddZnCl_2}=\dfrac{0,3.136.100\%}{336,557}=12,12\%\)

\(n_{H2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)

        1          2               1         1

       0,05    0,1           0,05      0,05 

    \(MgO+2HCl\rightarrow MgCl_2+H_2O|\)

       1           2              1            1

      0,2       0,4            0,2

a) \(n_{Mg}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)

\(m_{Mg}=0,05.24=1,2\left(g\right)\)

\(m_{MgO}=9,2-1,2=8\left(g\right)\)

⁰/₀Mg = \(\dfrac{1,2.100}{9,2}=13,04\)⁰/₀

⁰/₀MgO = \(\dfrac{8.100}{9,2}=86,96\)⁰/₀

b) Có : \(m_{MgO}=8\left(g\right)\)

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

\(n_{HCl\left(tổng\right)}=0,1+0,4=0,5\left(mol\right)\)

\(m_{HCl}=0,5.36,5=18,25\left(g\right)\)

\(m_{ddHCl}=\dfrac{18,25.100}{14,6}=125\left(g\right)\)

c) \(n_{MgCl2\left(tổng\right)}=0,05+0,2=0,25\left(mol\right)\)

⇒ \(m_{MgCl2}=0,25.95=23,75\left(g\right)\)

\(m_{ddspu}=9,2+125-\left(0,05.2\right)=134,1\left(g\right)\)

\(C_{MgCl2}=\dfrac{23,75.100}{134,1}=17,71\)⁰/₀

 Chúc bạn học tốt

a)\(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

PTHH: Mg + 2HCl → MgCl₂ + H₂

Mol:    0,05    0,1           0,05        0,05

PTHH: MgO + 2HCl → MgCl₂ + H₂O

Mol:     0,2         0,4         0,2

\(\%m_{Mg}=\dfrac{0,05.24.100\%}{9,2}=13,04\%;\%m_{MgO}=100-13,04=86,96\%\)

\(n_{MgO}=\dfrac{9,2-0,05.24}{40}=0,2\left(mol\right)\)

b,\(m_{HCl}=\left(0,1+0,4\right).36,5=18,25\left(g\right)\)

\(m_{ddHCl}=\dfrac{18,25.100}{14,6}=125\left(g\right)\)

c,m_{dd sau pứ} = 9,2+125-0,05.2 = 134,1 (g)

 \(C\%_{ddMgCl_2}=\dfrac{\left(0,05+0,2\right).95.100\%}{134,1}=17,71\%\)

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

\(MgO + 2HCl \rightarrow MgCl_2 + H_2O\)      (2)

Khí thu được là H₂

\(n_{H_2}= \dfrac{1,12}{22,4}=0,05 mol\)

Theo PTHH (1):

\(n_{Mg}= n_{H_2}= 0,05 mol\)

\(\Rightarrow m_{Mg}= 0,05 . 24= 1,2 g\)

\(\Rightarrow m_{MgO}= 9,2 - 1,2= 8g\)

C%_\(Mg\)= \(\dfrac{1,2}{9,2} .100\)%=13,04%

C%_\(MgO\)= 100% - 13,04%=86,96%

b)

\(n_{MgO}= \dfrac{8}{40}=0,2 mol\)

Theo PTHH (1) và (2):

\(n_{HCl(1)}= 2n_{Mg}= 0,1 mol\)

\(n_{HCl(2)}= 2n_{MgO}= 0,4 mol\)

Suy ra: \(n_{HCl}= n_{HCl(1)} + n_{HCl(2)}\)

                     \(= 0,1 + 0,4= 0,5 mol\)

\(\Rightarrow m_{HCl}= 0,5 . 36,5= 18,25g\)

\(\Rightarrow m_{dd HCl} = \dfrac{18,25 . 100%}{14,6%}=125 g\)

_c)

\(\)Dung dịch sau pư: MgCl₂

Theo PTHH:

\(n_{MgCl_2}= \dfrac{1}{2} n_{HCl}= 0,25 mol\)

\(\Rightarrow m_{MgCl_2}= 0,25 . 95=23,75g\)

\(m_{dd sau pư} = m_{Mg} + m_{MgO} + m_{dd HCl}- m_{H_2}\)

               \(= 9,2 + 125 + 2 . 0,05\)

             \(=134,1 g\)

C%_{\(MgCl_2\)}=\(\dfrac{23,75}{134,1}. 100\)%=17,71%

a) Gọi $n_{Al} = a(mol) ; n_{Fe} = b(mol) \Rightarrow 27a + 56b = 33,4(1)$

$2Al + 6HCl \to 2AlCl_3 + 3H_2$
$Fe + 2HCl \to FeCl_2 + H_2$
Theo PTHH : $n_{H_2} = 1,5a + b = \dfrac{17,92}{22,4} = 0,8(2)$

Từ (1)(2) suy ra : a = 0,2 ; b = 0,5

$\%m_{Al} = \dfrac{0,2.27}{33,4}.100\% = 16,17\%$
$\%m_{Fe} = 100\% - 16,17\% = 83,83\%$

b) $n_{HCl} = 2n_{H_2} = 1,6(mol)$

c) $m_{muối} = m_{hh} + m_{HCl} - m_{H_2} = 33,4 + 1,6.36,5 - 0,8.2 = 90,2(gam)$

PTHH: \(K_2O+2HCl\rightarrow2KCl+H_2O\)

            \(K_2SO_3+2HCl\rightarrow2KCl+SO_2\uparrow+H_2O\)

a) Ta có: \(\left\{{}\begin{matrix}n_{HCl}=\dfrac{200\cdot14,6\%}{36,5}=0,8\left(mol\right)\\n_{SO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{K_2SO_3}=0,3\left(mol\right)\\n_{K_2O}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{K_2O}=\dfrac{0,1\cdot94}{0,1\cdot94+0,3\cdot158}\cdot100\%\approx16,55\%\\\%m_{K_2SO_3}=83,45\%\end{matrix}\right.\)

b) Theo các PTHH: \(n_{KCl}=0,8\left(mol\right)\) \(\Rightarrow m_{KCl}=74,5\cdot0,8=59,6\left(g\right)\)

Mặt khác: \(\left\{{}\begin{matrix}m_{hh}=56,8\left(g\right)\\m_{SO_2}=0,3\cdot64=19,2\left(g\right)\end{matrix}\right.\)

\(\Rightarrow m_{dd}=m_{hh}+m_{ddHCl}-m_{SO_2}=237,6\left(g\right)\)

\(\Rightarrow C\%_{KCl}=\dfrac{59,6}{237,6}\cdot100\%\approx25,1\%\)

Câu 1:

Đặt \(n_{Fe}=x\left(mol\right);n_{Al}=y\left(mol\right)\)

\(n_{HCl}=0,4\left(mol\right)\)

\(Fe^o\rightarrow Fe^{+2}+2e\)

x_____________2x_(mol)

\(Al^o\rightarrow Al^{+3}+3e\)

y____________3y_(mol)

\(2H^-\rightarrow H_2^o+2e\)

0,8_____0,4____0,8_(mol)
\(BTe:2x+3y=0,8\)

Theo đề ta có hệ: \(\left\{{}\begin{matrix}56x+27y=11\\2x+3y=0,8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{Fe}=\frac{0,1.56}{11}.100\%=51\left(\%\right)\\\%m_{Al}=100-51=49\left(\%\right)\end{matrix}\right.\)

\(BTNT:\Rightarrow\left\{{}\begin{matrix}n_{FeCl_2}=0,1\left(mol\right)\\n_{AlCl_3}=0,2\left(mol\right)\end{matrix}\right.\)

\(m_{hh}=0,1.127+133,5.0,2=39,4\left(g\right)\)

\(m_{ddHCl}=\frac{36,5.0,8.100}{7,3}=400\left(g\right)\)

Câu 2:

\(n_{H_2}=0,15\left(mol\right)\)

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

(mol)____0,1____0,3______0,1______0,15__

\(\%m_{Al_2O_3}=\frac{7,8-27.0,1}{7,8}.100\%=65,4\left(\%\right)\)

Câu 3:

\(n_{H_2}=0,1\left(mol\right)\)

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

(mol)_____0,1__________________0,1__

\(\%m_{ZnO}=\frac{10,55-0,1.65}{10,55}.100\%=38,4\left(\%\right)\)