a) \(2Al+6HCl\rightarrow2AlCl_3+3H_2\left(1\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\left(2\right)\\ 2Al+2NaOH+2H_2O\rightarrow2NaAlO_2+3H_2\)

Cho hỗn hợp tác dụng với NaOH, chất rắn không tan là Fe

=> m_Fe= 1,12 (g) \(\Rightarrow n_{Fe}=0,02\left(mol\right)\)

Ta có: \(n_{H_2\left(2\right)}=n_{Fe}=0,02\left(mol\right)\)

=> \(n_{H_2\left(1\right)}=\Sigma n_{H_2}-n_{H_2\left(2\right)}=0,065-0,02=0,045\left(mol\right)\)

\(\Rightarrow n_{Al}=\dfrac{2}{3}n_{H_2\left(1\right)}=0,03\left(mol\right)\)

\(\Rightarrow m_{Al}=0,03.27=0,81\left(g\right)\)

\(\Rightarrow\%m_{Al}=41,97\%,\%m_{Fe}=58,03\%\)

b) \(m_{FeCl_2}=0,02.127=2,54\left(g\right)\\ m_{AlCl_3}=0,03.133,5=4,005\left(g\right)\)

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

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

\(AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3+3NaCl\)

\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaCl\)

\(NaOH+Al\left(OH\right)_3\rightarrow NaAlO_2+2H_2O\)

\(4Fe\left(OH\right)_2+O_2\underrightarrow{^{^{t^0}}}2Fe_2O_3+4H_2O\)

Ta có : 

\(n_{Fe_2O_3}=\dfrac{16}{160}=0.1\left(mol\right)\)

Dựa vào PTHH ta thấy : 

\(n_{Fe}=2\cdot n_{Fe_2O_3}=2\cdot0.1=0.2\left(mol\right)\)

\(m_{Fe}=0.2\cdot56=11.2\left(g\right)\)

\(\Rightarrow m_{Al}=19.3-11.2=8.1\left(g\right)\)

\(\%Al=\dfrac{8.1}{19.3}\cdot100\%=41.96\%\)

Bài 1:

Ta có: \(n_{H_2}=\dfrac{0,4}{2}=0,2\left(mol\right)\)

Bảo toàn Hidro: \(n_{HCl}=2n_{H_2}=0,4\left(mol\right)\) \(\Rightarrow m_{HCl\left(p.ứ\right)}=0,4\cdot36,5=14,6\left(g\right)\)

Bảo toàn khối lượng: \(m_{muối}=m_{KL}+m_{HCl\left(p.ứ\right)}-m_{H_2}=24,7\left(g\right)\)

Bài 2:

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

Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)=n_{Fe}\)

\(\Rightarrow\%m_{Fe}=\dfrac{0,1\cdot56}{12,8}\cdot100\%=43,75\%\) \(\Rightarrow\%m_{FeO}=56,25\%\)

Bài 3:

PTHH: \(KOH+HCl\rightarrow KCl+H_2O\)

Ta có: \(n_{KOH}=0,1\cdot1=0,1\left(mol\right)=n_{HCl}\)

*Bạn bổ sung thêm thể tích ddHCl thêm nhé !! 

B1:

2NaOH+H₂SO₄\(\rightarrow\)Na₂SO₄+2H₂O

n_NaOH=\(\frac{4}{40}=0.1\)mol

=>n_H2SO4=\(\frac{1}{2}\)n_NaOH=0.05 mol

=>C_M=\(\frac{n_{H2SO42}}{V}\)=\(\frac{0.05}{200}\)=2,5.10^-4 (M)

B2:

Mg+\(\frac{1}{2}\)O₂\(\underrightarrow{t^0}\)MgO                (1)

MgO+2HCl\(\rightarrow\)MgCl₂+H₂O (2)

n_Mg(1)=\(\frac{0,36}{24}=0,015mol\)

=>n_MgO(1)=0,015=n_MgO(2)

n_HCl(2)=2n_MgO(2)=0,03mol

=>C_M(HCl)=\(\frac{n_{HCl}}{V}=\frac{0,03}{100}=3.10^{-4}M\)

Vì Cu không tác dụng với HCl 

\(n_{H2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)

       1         2              1           1

      0,1     0,2                           0,1

a) \(n_{Fe}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

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

\(m_{Cu}=12-5,6=6,4\left(g\right)\)

b) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)

200ml = 0,2l

\(C_{M_{ddHCl}}=\dfrac{0,2}{0,2}=1\left(M\right)\)

c) ⁰/₀Fe = \(\dfrac{5,6.100}{12}=46,67\)⁰/₀

    ⁰/₀Cu = \(\dfrac{6,4.100}{12}=53,33\)⁰/₀

 Chúc bạn học tốt

a) $Fe + 2HCl \to FeCl_2 + H_2$

b) $n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)$

$n_{HCl} = \dfrac{200.7,3\%}{36,5} = 0,4(mol)$

$Fe + 2HCl \to FeCl_2 + H_2$

Ta thấy : $n_{Fe} : 1 = n_{HCl} : 2$ nên phản ứng vừa đủ

$n_{H_2} = n_{Fe} = 0,2(mol)$

$V_{H_2} = 0,2.22,4 = 4,48(lít)$

c) $m_{dd\ sau\ pư} = 11,2 + 200 - 0,2.2 = 210,8(gam)$

$C\%_{FeCl_2} = \dfrac{0,2.127}{210,8}.100\% = 12,05\%$

Không ai trả lời đâu :)))

\(BTNT\left(Mg\right):n_{Mg}=n_{MgO}=0,1mol\\ BTNT\left(Fe\right):n_{Fe}=2n_{Fe_2O_3}\\ \Rightarrow n_{Fe_2O_3}=0,2.2=0,4mol\\ BTNT\left(Al\right):n_{Al}=2n_{Al_2O_3}\\ \Rightarrow n_{Al_2O_3}=0,3.2=0,6mol\\ m_{rắn}=0,4.160+0,1.40+0,6.102=129,2g\)

\(BTNT\left(Mg\right):n_{Mg}=n_{MgO}=0,1mol\\ BTNT\left(Fe\right):n_{Fe}=2n_{Fe_2O_3}\\ \Rightarrow n_{Fe_2O_3}=0,2:2=0,1mol\\ BTNT\left(Al\right):n_{Al}=2n_{Al_2O_3}\\ \Rightarrow n_{Al_2O_3}=0,3:2=0,15mol\\ m_{rắn}=0,1.160+0,1.40+0,15.102=35,3g\)