2Al + 2H₂O + 2NaOH→ 3H₂ + 2NaAlO₂

0,2mol 0,3mol

mAl=0,2.27=5,4g

2Al + 6HCl→ 2AlCl₃+ 3H₂

0,2mol 0,3mol

Fe + 2HCl→ FeCl₂+ H₂

0,15mol 0,45-0,3 mol

mFe=0,15.56=8,4g

mCu=32,8-(6,4+8,4)=18g

%mFe=\(\frac{8,4}{32,8}.100=25,6\%\)

%mCu=\(\frac{18}{32,8}.100=54,8\%\)

%mAl=19,6%

nè bạn...... cái khúc cuối là y=0,2 mol đó, b hc chuyên Hóa à?

$n_{H_2}=\dfrac{15,68}{22,4}=0,7mol$

$PTHH :$

$Zn+2HCl\to ZnCl_2+H_2$

$Fe+2HCl\to FeCl_2+H_2$

a.Gọi $n_{Zn}=a;n_{Fe}=b(a,b>0)$

$\text{Ta có :}$

$m_{hh}=65a+56b=43,7g$

$n_{H_2}=a+b=0,7mol$

$\text{Ta có hpt :}$

\(\left\{{}\begin{matrix}65a+56b=43,7\\a+b=0,7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,5\\b=0,2\end{matrix}\right.\)

\(\Rightarrow\) $m_{Zn}=0,5.65=32,5g$

$m_{Fe}=43,7-32,5=11,2g$

$b.n_{Fe_3O_4}=\dfrac{46,4}{232}=0,2mol$

$Fe_3O_4+4H_2\overset{t^o}\to 3Fe+4H_2O$

$\text{Theo pt : 1 mol 4 mol}$

$\text{Theo đbài : 0,2 mol 0,7 mol}$

Tỷ lệ : $\dfrac{0,2}{1}>\dfrac{0,7}{4}$

$\text{=>Sau pư Fe3O4 dư}$

$\text{Theo pt :}$

$n_{Fe}=3/4.n_{H_2}=3/4.0,7=0,525mol

$=>m_{Fe}=0,525.56=29,4g$

Do Zn và Fe có số mol bằng nhau:

\(\rightarrow n_{Zn}=n_{Fe}=\dfrac{1}{2}=0,5\left(mol\right)\)

PTHH:

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

0,5                                   0,5

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

0,5                                 0,5

\(m_{Fe}=0,5.56=28\left(g\right)\)

\(m_{Zn}=0,5.65=32,5\left(g\right)\)

\(c,V_{H_2}=\left(0,5+0,5\right).24,79=24,79\left(l\right)\)

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

\(n_{H_2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\\ m_{HCl}=200.27,375\%=54,75\left(g\right)\\ n_{HCl}=\dfrac{54,75}{36,5}=1,5\left(mol\right)\)

PTHH:

Zn + 2HCl ---> ZnCl₂ + H₂

a ----> 2a --------> a -----> a

Fe + 2HCl ---> FeCl₂ + H₂

b ---> 2b -------> b ------> b

Hệ pt \(\left\{{}\begin{matrix}65a+56b=43,7\\a+b=0,7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,5\left(mol\right)\\b=0,2\left(mol\right)\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}m_{Zn}=0,5.65=32,5\left(g\right)\\m_{Fe}=0,2.56=11,2\left(g\right)\end{matrix}\right.\)

\(m_{dd}=43,7+200-0,7.2=242,3\left(g\right)\\ \rightarrow\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,5.136}{242,3}=28,06\%\\C\%_{FeCl_2}=\dfrac{0,2.127}{242,3}=10,48\%\\C\%_{HCl\left(dư\right)}=\dfrac{\left(1,5-0,5.2-0,2.2\right).36,5}{242,3}=1,51\%\end{matrix}\right.\)

\(n_{H_2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\\ pthh:\left\{{}\begin{matrix}Zn+H_2SO_4->ZnSO_4+H_2\\Fe+H_2SO_4->FeSO_{\text{ 4 }}+H_2\end{matrix}\right.\)
 gọi số mol Zn là x , số mol Fe là y 
=> 65x+56y=43,7
=> a+b=0,7 
=>a=0,5 , b =0,2  
=> \(m_{Zn}=0,5.65=32,5\\ m_{Fe}=43,7-32,5=11,2\left(G\right)\)
 

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

  giúp tôi với,cứu....!!!!

Ta có nH2 = 3,36/22,4 = 0,15 mol
        Fe +2 HCl -> FeCl2 + H2
       0,15.    0,3                   <-. 0,15.  ( Mol)
=> mFe = 0,15 × 56 = 8,4g
   => %Fe = 8,4/15×100% = 56% 
      => %Cu = 100% - 56% = 44%

=>VHCl =1\0,3=10\3 l

PTHH :  2Fe    +     6HCl   -->   2FeCl₃  +  3H₂   (1)

n_H2 =  \(\dfrac{3.36}{22.4}=0.15\left(mol\right)\)

Từ (1) ->  n_Fe = \(\dfrac{2}{3}n_{H_2}=0.1\left(mol\right)\)

-> m_Fe = n.M = 0,1  . 56 = 5.6 (g) => %m_Fe = \(\dfrac{5.6}{15}x100\%\approx37.3\%\)

-> %m_Cu =  100% - 37.3% = 62.7 % 

ai mà bt được

a: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

b: Đặt \(a=n_{Zn};b=n_{Fe}\)

Theo đề, ta có hệ phương trình:

\(\left\{{}\begin{matrix}65a+56b=37.2\\a+b=0.6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0.4\\b=0.2\end{matrix}\right.\)

\(m_{Zn}=0.4\cdot65=25\left(g\right)\)

\(m_{Fe}=0.5\cdot56=11.2\left(g\right)\)