\(a) n_{HCl} = \dfrac{200.14,6\%}{36,5} = 0,8(mol)\\ Fe + 2HCl \to FeCl_2 + H_2\\ n_{Fe} = n_{H_2} = \dfrac{1}{2}n_{HCl} = 0,4(mol)\\ \Rightarrow x = 0,4.56 = 22,4(gam)\\ b) V_{H_2} = 0,4.22,4 = 8,96(lít)\\ c) m_{dd\ sau\ pư} = 22,4 + 200 - 0,4.2 = 221,6(gam)\\ C\%_{FeCl_2} = \dfrac{0,4.127}{221,6}.100\% = 22,92\%\)

a) pthh:

1) Zn + 2HCl -> ZnCl2 + H2

2) ZnO + 2HCl -> ZnCl2 + H2O

b) ta có : V_H2 = 2,24l

=> n_H2 = 0,1mol

pt

1) Zn + 2HCl -> ZnCl2 + H2

1mol.....2mol......1mol.......1mol

0,1mol...0,2mol....0,1mol..0,1mol

=> n_Zn = 0,1mol

=> m_Zn = 0,1.65 = 6,5g

ta có m_Zn + m_ZnO = m_hh

=> m_ZnO = m_hh - m_Zn = 14,6 - 6,5 = 8,1g

c) ta có : n_HCl = 0,4mol

=> m_HCl = 0,4 . 36,5 = 14,6 g

=> m_ddHCl = 200g

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

x---------3x-----------x-------1,5x

Fe +2HCl----->FeCl2 +H2

y-------2y----------y------y

a)

n\(_{H2}=\)\(\frac{4,48}{22,4}=0,2mol\)

Theo bài ra ta có pt

\(\left\{{}\begin{matrix}27x+56y=5,5\\1,5x+y=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)

%m\(_{Al}=\frac{0,1.27}{5,5}.100\%=49,09\%\)

%m\(_{Fe}=100\%-49,09\%=50,91\%\)

b)Theo pthh

n\(_{HCl}=2n_{H2}=0,4\left(mol\right)\)

mddHCl =\(\frac{0,4.36,5.100}{14,6}=100\left(g\right)\)

mdd =5,5 + 100-0,4=105,1(g)

Theo pthh

n\(_{AlCl3}=n_{Al}=0,1mol\)

%m\(_{AlC_{ }l3}=\frac{0,1.98}{105,1}.100\%=9,32\%\)

Theo pthh

n\(_{FeCl2}=n_{Fe}=0,2mol\)

C%FeCl2 =\(\frac{0,2.56}{105,1}.100\%=10,66\%\)

Chúc bạn hok tốt

\(n_{Al}=x;n_{Fe}=y\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ hpt:\left\{{}\begin{matrix}27x+56y=5,5\\1,5x+y=\frac{4,48}{22,4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\\\rightarrow\left\{{}\begin{matrix}\%m_{Al}=\frac{0,1.27}{5,5}.100\%=49,1\left(\%\right)\\\%m_{Fe}=100-49,1=50,9\left(\%\right)\end{matrix}\right.\\ m_{ddHCl}=\frac{100.\left[36,5.\left(3x+2y\right)\right]}{14,6}=100\left(g\right)\\ C\%_M=\frac{0,1.133,5+127.0,05}{5,5+100-2.\left(1,5x+y\right)}.100\%=18,74\left(\%\right)\)

hình như đề sai

Mg + 2HCl -> MgCl2 + H2

Fe + 2HCl -> FeCl2 + H2

ⁿH2 = 3,36/22,4 = 0,15 (mol)

Tổng ⁿHCl = 2ⁿH2 = 2.0,15 = 0,3 (mol)

^VHCl = 0,3.22,4 = 6,72 (l)

Mg + 2HCl → MgCl₂ + H₂↑ (1)

Fe + 2HCl → FeCl₂ + H₂↑ (2)

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

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

ta có: \(\left\{{}\begin{matrix}24x+56y=5,2\\x+y=0,15\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)

Vậy \(n_{Mg}=0,1\left(mol\right)\)

\(n_{Fe}=0,05\left(mol\right)\)

theo PT1: \(n_{HCl}=2n_{Mg}=2\times0,1=0,2\left(mol\right)\)

theo PT2: \(n_{HCl}=2n_{Fe}=2\times0,05=0,1\left(mol\right)\)

\(\Rightarrow\Sigma n_{HCl}=0,2+0,1=0,3\left(mol\right)\)

\(\Rightarrow V_{ddHCl}=\dfrac{0,3}{1}=0,3\left(l\right)=300\left(ml\right)\)

ta có: n_H2= \(\dfrac{3,36}{22,4}\)= 0,15( mol)

PTPU

Mg+ 2HCl\(\rightarrow\) MgCl₂+ H₂\(\uparrow\)

Fe+ 2HCl\(\rightarrow\) FeCl₂+ H₂\(\uparrow\)

theo các ptpu có: n_HCl= 2n_H2= 0,3( mol)

\(\Rightarrow\) V_{dd HCl}= \(\dfrac{0,3}{1}\)= 0,3( lít)

Bạn ơi đây là môn toán chứ có phải môn hóa đâu

\(a.n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\\ \left[Na^+\right]=\left[OH^-\right]=\left[NaOH\right]=\dfrac{0,5}{0,5}=1\left(M\right)\\ b.HCl+NaOH\rightarrow NaCl+H_2O\\ n_{HCl}=n_{NaOH}=0,5\left(mol\right)\\ V_{ddHCl}=\dfrac{0,5}{2}=0,25\left(l\right)\)