Gọi x, y, z lần lượt là số mol của Mg, Fe và Zn.

Ta có: \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

Theo đề, ta có: 

24x + 56y + 65z = 21 (*)

\(44,8x=22,4z\) (**)

PTHH:

\(Mg+H_2SO_4--->MgSO_4+H_2\uparrow\left(1_{ }\right)\)

\(Fe+H_2SO_4--->FeSO_4+H_2\uparrow\left(2\right)\)

\(Zn+H_2SO_4--->ZnSO_4+H_2\uparrow\left(3\right)\)

Từ PT_(1,2,3), ta có phương trình: x + y + z = 0,4 (***)

Từ (*), (**) và (***), ta có HPT:

\(\left\{{}\begin{matrix}24x+56y+65z=21\\44,8x=22,4z\\x+y+z=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\\z=0,2\end{matrix}\right.\)

\(\Rightarrow m_{Mg}=0,1.24=2,4\left(g\right)\)

\(m_{Fe}=56.0,1=5,6\left(g\right)\)

\(m_{Zn}=0,2.65=13\left(g\right)\)

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right);n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{H_2\left(tổng\right)}=n_{Fe}+n_{Zn}=0,2+0,2=0,4\left(mol\right)\\ n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right);n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,4.22,4=8,96\left(l\right)\\ m_{ZnCl_2}=0,2.136=27,2\left(g\right)\\ m_{FeCl_2}=127.0,2=25,4\left(g\right)\)

fe+2hcl---->fecl2+ h2

zn+2hcl-----> zncl2+ h2

goi so mol cua fe va zn lan luot la x, y(x,y>0)

ta co he pt

(1) 56x+65y=12.1

(2) x+y=4.48/22.4=0.2

rui ban giai ra la duoc

b> xet 2pt thay n HCL= 2nH2===> mHCL= 2*0.2*3605=.....

\(n_{Mg}=a;n_{Fe}=0,5a;n_{Zn}=b\\ a\left(24+28\right)+65b=52a+65b=44,2\\ 1,5a+b=\dfrac{24,64}{22,4}1,1\\ a=0,6;b=0,2\\ \%m_{Mg}=\dfrac{24a}{44,2}=32,58\%\\ \%m_{Fe}=\dfrac{28a}{44,2}=38\%\\ \%m_{Zn}=29,42\%\\ m_{ddacid}=\dfrac{98\left(1,5a+b\right)}{0,08}=1347,5g\\ m_{ddsau}=1389,5g\\ C\%_{MgCl_2}=\dfrac{95a}{1389,5}=4,10\%\\ C\%_{FeCl_2}=\dfrac{127.0,5a}{1389,5}=2,74\%\\ C\%_{ZnCl_2}=\dfrac{136b}{1389,5}=1,96\%\)

  giúp tôi với,cứu....!!!!

a, Fe + 2HCl ---> FeCl₂ + H₂ (1)

Mg + 2HCl ---> MgCl₂ + H₂ (2)

b, Số mol H₂ là:

n= \(\frac{V}{22,4}\)= \(\frac{4,48}{22,4}\)= 0,2 ( mol )

Đã lm lại còn k lm cho hết

nHCl=19,04/22,4=0,85(mol)

Mg+2HCl--->MgCl2+H2

x_____2x_____x_____x

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

y____3y_____y_______3/2y

Zn+2HCl--->ZnCl2+H2

1/2y__1/4y___1/2y___1/2y

Hệ pt:

\(\left\{{}\begin{matrix}24x+92y=35\\x+2y=0,85\end{matrix}\right.\)

=>x,y

=>mAl,Fe,Mg