PTHH: 2Al + 6HCl ===> 2AlCl3 + 3H2

Fe + 2HCl ===> FeCl2 + H2

a/ Gọi số mol Al, Fe lần lượt là x, y (mol)

nH2 = 5.6 / 22,4 = 0,25 mol

Theo đề ra, ta có hệ phương rình sau:

{27x+56y=8,4

1,5x+y=0,25

=> {x=0,1==>mAl=0,1.27=2,7 g

y=0,1==>mFe=0,1.56=5,6 g

=> %mAl = 2,7\8,3.100%=32,5%

=> %mFe = 100% - 32,5% = 67,5%

b/ Theo phương trình, ta thấy :

∑nHCl=0,2+0,3=0,5(mol)

=> VHCl = 0,5 / 2 = 0,25 lít

a,

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

x_______3x______x______x

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

y_______2y_______y____y

Giải hệ PT:

\(\left\{{}\begin{matrix}27x+56y=8,3\\\frac{3}{2}x+y=0,25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)

b,\(m_{Al}=0,1.27=2,7\left(g\right)\)

\(m_{Fe}=0,1.56=5,6\left(g\right)\)

c,\(\%_{Al}=\frac{2,7}{8,3}.100\%=32,53\%\)

\(\%_{Fe}=100\%-32,53\%=67,47\%\)

d,\(n_{HCl}=3.0,1+2.0,1=0,5\left(mol\right)\)

\(\Rightarrow V_{HCl}=\frac{0,5}{2}=0,25\left(l\right)=250\left(ml\right)\)

e,\(CM_{AlCl3}=\frac{0,1}{0,25}=0,4M\)

\(CM_{FeCl2}=\frac{0,1}{0,25}=0,4M\)

Em đăng sang môn Hoá nha

\(2Al+6HCl->2AlCl_3+3H_2\\ Fe+2HCl->FeCl_2+H_2\\ n_{Al}=a;n_{Fe}=b\\ 27a+56b=8,3\\ 1,5a+b=\dfrac{5,6}{22,4}=0,25\\ a=b=0,1\\ m_{Al}=27\cdot0,1=2,7g\\ m_{Fe}=8,3-2,7=5,6g\\ a=\dfrac{3a+2b}{500}\cdot36,5=3,65\%\\ m_{ddsau}=508,3-0,25\cdot2=507,8g\\ C\%_{AlCl_3}=\dfrac{133,5a}{507,8}=2,63\%\\ C\%_{FeCl_2}=\dfrac{127b}{507,8}=2,50\%\)

Mong MN giúp mình nhanh với , mình đang rất gấp 

Cảm ơn mọi người nhiều nhà 😘😘

a)

Gọi $n_{Fe} = a ; n_{FeO} = b; n_{FeCO_3} = c \Rightarrow 56a + 72b + 116c = 21,6(1)$

$Fe + 2HCl \to FeCl_2 + H_2$
$FeO + 2HCl \to FeCl_2 + H_2O$
$FeCO_3 + 2HCl \to FeCl_2 + CO_2 + H_2O$

\(\dfrac{2a+44c}{a+c}=15.2=30\left(2\right)\)

$n_{FeCl_2} = a + b + c= \dfrac{31,75}{127} = 0,25(3)$
Từ (1)(2)(3) suy ra a = 0,05 ; b = 0,1 ; c = 0,1

$n_{HCl} = 2a + 2b + 2c =0,5(mol)$
$m_{dd\ HCl} = \dfrac{0,5.36,5}{7,3\%} = 250(gam)$

$\%m_{Fe} = \dfrac{0,05.56}{21,6}.100\% = 12,96\%$
$\%m_{FeO} = \dfrac{0,1.72}{21,6}.100\% = 33,33\%$

$\%m_{FeCO_3} = 53,71\%$

\(a.Fe+2HCl\rightarrow FeCl_2+H_2\\b.n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ n_{H_2}=n_{Fe}=0,1\left(mol\right)\\ \Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\\ c.n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\\ m_{FeCl_2}=0,1.127=12,7\left(g\right) \)

Fe + 2HCl -> FeCl2 + H2
nFe = 5,6/56 = 0,1 mol
=>nH2 = 0,1 mol
=> VH2= 0,1*22,4= 2,24 lít

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PTHH: Fe + 2HCl ---> FeCl₂ + H₂

          0,1-->0,2------------------>0,1

=> \(\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{0,2.36,5}{15\%}=\dfrac{146}{3}\left(g\right)\\V_{H_2}=0,1.22,4=4,48\left(l\right)\end{matrix}\right.\)

Đáp án A

a)

$Fe + 2HCl \to FeCl_2 + H_2$
$FeO + 2HCl \to FeCl_2 + H_2O$

b)

Theo PTHH : $n_{Fe} = n_{H_2} = \dfrac{3,36}{22,4} = 0,15(mol)$
$m_{Fe} = 0,15.56 = 8,4(gam)$
$m_{FeO} = 12 - 8,4 = 3,6(gam)$

$n_{FeO} =0,05(mol)$
Theo PTHH : $n_{HCl} = 2n_{Fe} + 2n_{FeO} = 0,4(mol)$

$V_{dd\ HCl} = \dfrac{0,4}{2} = 0,2(lít)$

c) $Fe + CuSO_4 \to FeSO_4 + Cu$

$n_{Cu} = n_{Fe} = 0,15(mol) \Rightarrow m_{chất\ rắn} = m_{FeO} + m_{Cu}$

$= 3,6 + 0,15.64 = 13,2(gam)$