Sửa: $V_{H_2}=7,168(l)$

$a\bigg)$

Đặt $n_{Mg}=x;n_{Fe}=y;n_{Al}=z$

$\to 24x+56y+27z=9,52(1)$

$n_{H_2}=\dfrac{7,168}{22,4}=0,32(mol)$

$n_{Cl_2}=\dfrac{8,064}{22,4}=0,36(mol)$

BTe: $x+y+1,5z=n_{H_2}=0,32(2)$

BTe: $x+1,5y+1,5z=n_{Cl_2}=0,36(3)$

Từ $(1)(2)(3)\to x=0,12(mol);y=0,08(mol);z=0,08(mol)$

$\to \begin{cases} \%m_{Mg}=\dfrac{0,12.24}{9,52}.100\%=30,25\%\\ \%m_{Fe}=\dfrac{0,08.56}{9,52}.100\%=47,06\%\\ \%m_{Al}=100-47,06-30,25=22,69\% \end{cases}$

$b\bigg)$

Bảo toàn H: $n_{HCl}=2n_{H_2}=0,64(mol)$

$\to C_{M_{HCl}}=\dfrac{0,64}{0,2}=3,2M$

$\to a=3,2$

$c\bigg)$

Dung dịch sau gồm $MgCl_2,FeCl_2,AlCl_3$

Bảo toàn $Mg,Al,Fe:n_{MgCl_2}=0,12(mol);n_{AlCl_3}=n_{FeCl_2}=0,08(mol)$

$\to C_{M_{MgCl_2}}=\dfrac{0,12}{0,2}=0,6M$

$\to C_{M_{AlCl_3}}=C_{M_{FeCl_2}}=\dfrac{0,08}{0,2}=0,4M$

$a\bigg)$

Đặt $n_{Mg}=x;n_{Fe}=y;n_{Al}=z$

$\to 24x+56y+27z=9,52(1)$

$n_{H_2}=\dfrac{14,336}{22,4}=0,64(mol)$

$n_{Cl_2}=\dfrac{8,064}{22,4}=0,36(mol)$

BTe: $x+y+1,5z=n_{H_2}=0,64(2)$

BTe: $x+1,5y+1,5z=n_{Cl_2}=0,36(3)$

Từ $(1)(2)(3)\to$ nghiệm âm, xem lại đề

giúp em vs

a)

Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\)

=> 56a + 24b = 18,4 (1)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

             a-->2a------>a------>a

             Mg + 2HCl --> MgCl₂ + H₂

             b--->2b------->b------>b

=> \(a+b=\dfrac{11,2}{22,4}=0,5\) (2)

(1)(2) => a = 0,2 (mol); b = 0,3 (mol)

\(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,2.56}{18,4}.100\%=60,87\%\\\%m_{Mg}=\dfrac{0,3.24}{18,4}.100\%=39,13\%\end{matrix}\right.\)

b) \(n_{HCl\left(pư\right)}=2a+2b=1\left(mol\right)\)

=> \(n_{HCl\left(tt\right)}=\dfrac{1.125}{100}=1,25\left(mol\right)\)

=> m_HCl(tt) = 1,25.36,5 = 45,625 (g)

=> \(a=\dfrac{45,625.100}{18,25}=250\left(g\right)\)

c) 

m_{dd sau pư} = 18,4 + 250 - 0,5.2 = 267,4 (g)

\(C\%_{FeCl_2}=\dfrac{0,2.127}{267,4}.100\%=9,5\%\) 

\(C\%_{MgCl_2}=\dfrac{0,3.95}{267,4}.100\%=10,66\%\)

Mg+2HCl->MgCl2+H2

x------2x--------x---------x

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

y---------3y-----y--------3\2y

ta có :

\(\left\{{}\begin{matrix}24x+27y=11,7\\x+\dfrac{3}{2}y=0,6\end{matrix}\right.\)

=>x=0,15 mol, y=0,3 mol

=>%mMg=\(\dfrac{0,15.24}{11,7}.100=30,77\%\)

=>%mAl=100-30,77=69,23%

b)

m HCl=1,2.36,5=43,8g

=>C%=\(\dfrac{43,8}{200}.100\)=21,9%

$a)$

Đặt $n_{Al}=x(mol);n_{Fe}=y(mol)$

$\to 27x+56y=13,75(1)$

Bảo toàn e: $1,5x+y=n_{H_2}=\dfrac{11,2}{22,4}=0,5(2)$

Từ $(1)(2)\to x=0,25(mol);y=0,125(mol)$

$\to \%m_{Al}=\dfrac{0,25.27}{13,75}.100\%\approx 49,09\%$

$\to \%m_{Fe}=100-49,09=50,91\%$

$b)$

Bảo toàn H: $n_{HCl}=2n_{H_2}=1(mol)$

$\to a=\dfrac{1.36,5.120\%}{18,25\%}=240(g)$

$c)$

Bảo toàn Al,Fe: $n_{AlCl_3}=0,25(mol);n_{FeCl_2}=0,125(mol)$

$m_{dd_{HCl(p/ứ)}}=\dfrac{1.36,5}{18,25\%}=200(g)$

Ta có $m_{dd\, sau}=13,75+200-0,5.2=212,75(g)$

$\to \begin{cases} C\%_{AlCl_3}=\dfrac{0,25.133,5}{212,75}.100\%=15,69\%\\ C\%_{FeCl_2}=\dfrac{0,125.127}{212,75}.100\%=7,46\% \end{cases}$

thank anh/chị nhé

Đáp án B

\(1)n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ Mg+2HCl\to MgCl_2+H_2\\ Fe+2HCl\to FeCl_2+H_2\)

Từ giả thiết và theo PT: 

\(\begin{cases} 24n_{Mg}+56n_{Fe}=5,2\\ n_{Mg}+n_{Fe}=0,15 \end{cases}\\ \Rightarrow n_{Mg}=0,1(mol);n_{Fe}=0,05(mol)\)

\(\Rightarrow \begin{cases} \%m_{Mg}=\dfrac{0,1.24}{5,2}.100\%=46,15\%\\ \%m_{Fe}=100-46,15=53,85\% \end{cases}\\ 2)\Sigma n_{HCl}=2n_{H_2}=0,3(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,3}{1}=0,3(l)=300(ml)\)

lười làm thì đừng làm

box hóa có luật không tham khảo rồi

Đáp án C

Gọi số mol Mg, Fe, Al là a, b, c

=> 24a + 56b + 27c = 23,8

PTHH: Mg + 2HCl --> MgCl₂ + H₂ 

            a------------------------->a

            Fe + 2HCl --> FeCl₂ + H₂

            b------------------------->b

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

            c------------------------->1,5c

=> a + b + 1,5c = \(\dfrac{17,92}{22,4}=0,8\left(mol\right)\)

PTHH: Mg + Cl₂ --t^o--> MgCl₂

             a-->a

            2Fe + 3Cl₂ --t^o--> 2FeCl₃

             b--->1,5b

             2Al + 3Cl₂ --t^o--> 2AlCl₃

             c--->1,5c

=> \(a+1,5b+1,5c=\dfrac{20,16}{22,4}=0,9\left(mol\right)\)

=> a = 0,3; b = 0,2; c = 0,2

=> \(\left\{{}\begin{matrix}m_{Mg}=0,3.24=7,2\left(g\right)\\m_{Fe}=0,2.56=11,2\left(g\right)\\m_{Al}=0,2.27=5,4\left(g\right)\end{matrix}\right.\)