a. Cu+Cl₂-> CuCl₂

    1     1       1 (mol)

 0,2    0,2      0,2(mol)

n_Cu=m/M=12,8:64=0,2 mol

b. V_Cl2= n.22,4= 0,2.22,4= 4,48 (l)

c. m_Cl2= n.M=0,2. 71= 14,2 g

K nha

a, PTHH:\(Cu+Cl_2\rightarrow CuCl_2\)

b, \(n_{Cu}=\frac{m}{M}=\frac{12,8}{64}=0,2\left(mol\right)\)

Ta thấy \(n_{CuCl_2}=n_{Cl_2}=n_{Cu}=0,2\left(mol\right)\)

\(V_{Cl_2}=n_{Cl_2}.22,4=0,2.22,4=4,48\left(l\right)\)

c, \(m_{CuCl_2}=n_{CuCl_2}.M=0,2.135=27\left(g\right)\)

Đáp án là B. 73,5%.

a)

\(n_{Cl_2} = \dfrac{6,72}{22,4}\ =0,3(mol)\)

m_muối= m_{kim loại} + m_Cl2 = 8,3 + 0,3.71 = 29,6(gam)

b)

\(\left\{{}\begin{matrix}Al:x\left(mol\right)\\Fe:y\left(mol\right)\end{matrix}\right.\)⇒27x + 56y = 8,3(1)

\(2Al + 3Cl_2 \xrightarrow{t^o} 2AlCl_3\\ 2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3\)

Suy ra : 1,5x + 1,5y = 0,3(2)

Từ (1)(2) suy ra : x = 0,1 ; y = 0,1

Vậy :

\(\%m_{Al} = \dfrac{0,1.27}{8,3}.100\% = 32,53\%\\ \%m_{Fe} = 100\% - 32,53\% = 67,47\%\)

\(Đặt:n_{Al}=x\left(mol\right),n_{Fe}=y\left(mol\right)\)

\(m_{hh}=27x+56y=8.3\left(g\right)\left(1\right)\)

\(n_{Cl_2}=0.3\left(mol\right)\)

\(2Al+3Cl_2\underrightarrow{t^0}2AlCl_3\)

\(2Fe+3Cl_2\underrightarrow{t^0}2FeCl_3\)

\(n_{Cl_2}=1.5x+1.5y=0.3\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):x=y=0.1\)

\(m_{Muối}=m_{AlCl_3}+m_{FeCl_3}=0.1\cdot133.5+0.1\cdot162.5=29.6\left(g\right)\)

\(\%Al=\dfrac{2.7}{8.3}\cdot100\%=32.53\%\)

\(\%Fe=100-32.53=67.47\%\)

\(PTHH:2Al+3Cl_2\) → \(2AlCl_3\)

\(n_{Al}=\dfrac{m}{M}=\dfrac{1,35}{27}=0,05\left(mol\right)\)

Theo PTHH

⇒ \(n_{Cl_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.0,05=0,075\left(mol\right)\)

⇒ \(V_{Cl_2\left(đktc\right)}=n.22,4=0,075.22,4=1,68\left(l\right)\)

n Mg = 3,6/24 = 0,15(mol)

n Al = 5,4/27 = 0,2(mol)

Gọi n Cl2 = a(mol) ; n O2 = b(mol)

Bảo toàn e : 

2n Cl2 + 4n O2 = 2n Mg + 3n Al

<=> 2a + 4b = 0,15.2 + 0,2.3 = 0,9(1)

Bảo toàn khối lượng : 

m X + m Al + m Mg = m Z

=> 71a + 32b = 29,95 - 3,6 -5,4 = 20,95(2)

Từ (1)(2) suy ra  a = 0,25 ; b = 0,1

%m O2 = 0,1.32/(0,25.71 + 0,1.32)   .100% = 15,27%