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__

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

Đề câu 2 sao khỏi làm đi

a)Gọi x,y lần lượt là số mol của Al, Fe trong hỗn hợp ban đầu (x,y>0)

Sau phản ứng hỗn hợp muối khan gồm: \(\left\{{}\begin{matrix}AlCl_3:x\left(mol\right)\\FeCl_2:y\left(mol\right)\end{matrix}\right.\)

Ta có hệ phương trình: \(\left\{{}\begin{matrix}27x+56y=13,9\\133,5x+127y=38\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\approx0,0896\\y\approx0,205\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,0896\cdot27\cdot100\%}{13,9}\approx17,4\%\\\%m_{Fe}=\dfrac{0,205\cdot56\cdot100\%}{13,9}\approx82,6\%\end{matrix}\right.\)

Theo Bảo toàn nguyên tố Cl, H ta có:\(n_{H_2}=\dfrac{n_{HCl}}{2}=\dfrac{3n_{AlCl_3}+2n_{FeCl_2}}{2}\\ =\dfrac{3\cdot0,0896+2\cdot0,205}{2}=0,3394mol\\ \Rightarrow V_{H_2}=0,3394\cdot22,4\approx7,6l\)

\(2Al\left(a\right)+6HCl\rightarrow2AlCl_3+3H_2\left(1,5a\right)\)

\(Al_2O_3\left(b\right)+6HCl\left(6b\right)\rightarrow2AlCl_3+3H_2O\)

\(1,5a=\dfrac{13,44}{22,4}=0,6\Rightarrow a=0,4\left(I\right)\)

\(27a+102b=21\left(II\right)\)

Thay (I) vào (II) => b = 0,1

\(\Rightarrow\%m_{Al}=\dfrac{27.0,4}{21}.100=51,43\%\Rightarrow\%m_{Al_2O_3}=48,57\%\)

c) \(m_{HCl}=6.0,1.36,5=21,9g\)

\(\Rightarrow m_{ddHCl}=\dfrac{100.21,9}{7,3}=300g\)

\(\Rightarrow V_{ddHCl}=\dfrac{300}{1,03}\approx291,26ml\approx0,3l\)

Giải:

a) Số mol H2 ở đktc là:

nH2 = V/22,4 = 13,44/22,4 = 0,6 (mol)

PTHH: 2Al + 6HCl -> 2AlCl3 + 3H2↑

-----------0,4----1,2------0,4---------0,6--

PTHH: Al2O3 + 6HCl -> 2AlCl3 + 3H2O

----------0,1---------0,6-----0,2---------0,3--

b) Thành phần phần trăm khối lượng mỗi chất trong hh ban đầu là:

%mAl = (mAl/mhh).100 = (0,4.27/21).100 ≃ 51,4 %

=> %mAl2O3 = 100 - 51,4 = 48,6 %

=> mAl2O3 = (21/100).48,6 = 10,2 (g)

=> nAl2O3 = m/M = 10,2/102 = 0,1 (mol)

c) Khối lượng dd HCl tham gia pư là:

mddHCl = (mct.100)/C% = (1,8.36,5.100)/7,3 = 900 (g)

Thể tích HCl đã dùng là:

VHCl = m/D = 900/1,03 ≃ 874 (ml)

Vậy ...

\(a)2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ b)n_{H_2}=\dfrac{5,6}{22,4}=0,25mol\\ n_{Al}=a;n_{Fe}=b\\ \left\{{}\begin{matrix}3a+b=0,25\\27a+56b=8,3\end{matrix}\right.\\ a=\dfrac{19}{470};b=\dfrac{121}{940}\\ \%m_{Al}=\dfrac{\dfrac{19}{470}\cdot27}{8,3}\cdot100=13,15\%\\ \%m_{Fe}=100-13,15=86,85\%\\ c)n_{HCl}=3\cdot\dfrac{19}{470}+2\cdot\dfrac{121}{940}=\dfrac{89}{235}mol\\ m_{ddHCl=}=\dfrac{\dfrac{89}{235}\cdot36,5}{7,3}\cdot100=189g\\ d)n_{AlCl_3}=n_{Al}=\dfrac{19}{470}mol\\ n_{Fe}=n_{FeCl_2}=\dfrac{121}{940}mol\)

\(m_{dd}=8,3+189-0,25.2=196,8g\\ C_{\%AlCl_3}=\dfrac{\dfrac{19}{470}\cdot133,8}{196,8}\cdot100=2,8\%\\ C_{\%FeCl_2}=\dfrac{\dfrac{121}{940}127}{196,8}\cdot100=8,3\%\)

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)

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

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%

Tham khảo