a,\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH: Fe + 2HCl → FeCl₂ + H₂

Mol:     0,1    0,2                             0,1

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

\(\Rightarrow\%m_{Fe}=\dfrac{5,6.100\%}{12}=46,67\%;\%m_{Cu}=100-46,67=53,33\%\)

c,\(m_{HCl}=0,2.36,5=7,3\left(g\right)\Rightarrow m_{ddHCl}=\dfrac{7,3.100}{14,6}=50\left(g\right)\)

a) nH2=0,2(mol)

PTHH: Mg + 2 HCl -> MgCl2 + H2

0,2______0,4_______0,2_____0,2(mol)

PTHH: MgO +2 HCl -> MgCl2 + H2O

b) mMgO= 8,8 - 0,2.24=4(g)

%mMgO= (4/8,8).100= 45,455%

=>%Mg=54,545%

c) nHCl(tổng)= 2. nMg + 2. nMgO= 2. 0,2+ 0,1.2=0,6(mol)

=> mHCl= 0,6.36,5=21,9(g)

=>mddHCl=(21,9.100)/7,3=300(g)

d) mddMgCl2= mddHCl + m(hỗn hợp ban đầu) - mH2

<=>mddHCl= 300+ 8,8- 0,2.2= 308,4(g)

nMgCl2=0,3(mol) => mMgCl2= 0,3.95=28,5(g)

=>C%ddMgCl2= (28,5/308,4).100=9,241%

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) $Fe + CuSO_4 \to FeSO_4 + Cu$

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

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

a/ Fe + 2HCl \(\rightarrow\) FeCl₂ + H₂

n_H2 = \(\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Theo PTHH: n_H2 = n_Fe = 0,15 (mol) \(\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)

\(\Rightarrow m_{Cu}=11-8,4=2,6\left(g\right)\)

\(\Rightarrow\%m_{Fe}=\dfrac{8,4}{11}.100\%\approx76,4\%\)

\(\Rightarrow\%m_{Cu}=100-76,4\approx23,6\%\)

b/ Theo PTHH ta có: n_HCl = 2n_Fe = 2.0,15 = 0,3 (mol)

\(\Rightarrow V_{ddHCl}=\dfrac{0,3}{2}=0,15\left(M\right)\)

c/ m_HCl = 36,5 . 0,3 = 10,95(g)

\(\Rightarrow C\%_{HCl}=\dfrac{m_{HCl}}{m_{ddHCl}}.100\%=\dfrac{10,95}{200}.100\%=5,475\%\)

a, PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Theo PT: \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,15.56}{11}.100\%\approx76,36\%\\\%m_{Cu}\approx23,64\%\end{matrix}\right.\)

b, Theo PT: \(n_{HCl}=2n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow V_{ddHCl}=\dfrac{0,3}{2}=0,15\left(l\right)\)

c, \(C\%_{HCl}=\dfrac{0,3.36,5}{200}.100\%=5,475\%\)

Bạn xem lại xem đề cho hỗn hợp gồm gì nhé, nếu là hh CaO và Fe₂O₃ thì pư với dd HCl không thu được CO₂ đâu.

Cảm ơn ạ

Gọi a, b lần lượt là mol của Al và Zn

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

a                                        1,5a

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

b                                       b

\(\Rightarrow\left\{{}\begin{matrix}27a+65b=9,2\\1,5a+b=\dfrac{5,6}{22,4}\end{matrix}\right.\)                \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)

\(\%m_{Al}=\dfrac{0,1.27}{9,2}.100\%=29,35\%\)

\(\%m_{Zn}=\dfrac{0,1.65}{9,2}.100\%=70,35\%\)

b. \(n_{H_2}=0,25mol\)           \(\Rightarrow n_{HCl}=0,5mol\)

\(\Rightarrow m_{HCl}=0,5.36,5=18,25g\)

Ta có:  \(10\%=\dfrac{18,25}{m_{dd}}.100\%\)

\(\Leftrightarrow m_{dd}=182,5g\)

\(a)n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{Zn}=n_{H_2}=n_{ZnCl_2}=0,1mol\\ m_{Zn}=0,1.65=6,5g\\ m_{Cu}=9,7-6,5=3,2g\\ b)C_{\%ZnCl_2}=\dfrac{0,1.136}{6,5+120-0,1.2}\cdot100=10,77\%\)