a, \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

\(FeCl_3+3KOH\rightarrow3KCl+Fe\left(OH\right)_{3\downarrow}\)

\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

Theo PT: \(n_{HCl}=6n_{Fe_2O_3}=0,6\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,6.36,5=21,9\left(g\right)\)

b, \(n_{Fe\left(OH\right)_3}=n_{FeCl_3}=2n_{Fe_2O_3}=0,2\left(mol\right)\)

\(\Rightarrow m_{Fe\left(OH\right)_3}=0,2.107=21,4\left(g\right)\)

\(n_{KOH}=3n_{FeCl_3}=0,6\left(mol\right)\)

\(\Rightarrow C_{M_{KOH}}=\dfrac{0,6}{0,2}=3\left(M\right)\)

a)

- Xét phần 1:

\(n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\)

PTHH: CaCO₃ + 2CH₃COOH --> (CH₃COO)₂Ca + CO₂ + H₂O

               0,1----->0,2

=> \(n_{CH_3COOH\left(P1\right)}=0,2\left(mol\right)\)

- Xét phần 2:

\(n_{H_2}=\dfrac{0,224}{22,4}=0,01\left(mol\right)\)

PTHH: Mg + 2CH₃COOH --> (CH₃COO)₂Mg + H₂

                           0,02<---------------------0,01

=> \(n_{CH_3COOH\left(P2\right)}=0,02\left(mol\right)\)

=> \(n_{CH_3COOH\left(tổng\right)}=0,2+0,02=0,22\left(mol\right)\)

=> \(m_{CH_3COOH\left(tổng\right)}=60.0,22=13,2\left(g\right)\)

b) 

\(n_{CH_3COOH\left(pư\right)}=\dfrac{0,22.75}{100}=0,165\left(mol\right)\)

PTHH: CH₃COOH + C₂H₅OH --H₂SO_4(đ)t^o--> CH₃COOC₂H₅ + H₂O

                 0,165------------------------------>0,165

=> \(m_{CH_3COOC_2H_5}=0,165.88=14,52\left(g\right)\)

c)

\(n_{H_2}=\dfrac{34,944}{22,4}=1,56\left(mol\right)\)

PTHH: 2CH₃COOH + 2Na --> 2CH₃COONa + H₂

                 0,22------------------------------->0,11

           2H₂O + 2Na --> 2NaOH + H₂

             2,9<------------------1,45

=> m_dd = 13,2 + 2,9.18 = 65,4 (g)

=> \(V=\dfrac{65,4}{1,02}=\dfrac{1090}{17}\left(ml\right)\)

\(x=C_{M\left(A\right)}=\dfrac{0,22}{\dfrac{1,09}{17}}=\dfrac{374}{109}M\)

1)

a, \(n_{Al}=\dfrac{15,3}{102}=0,15\left(mol\right)\)

PTHH: Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O

Mol:     0,15         0,9        0,3      

\(m_{ddHCl}=\dfrac{0,9.36,5.100}{20}=164,25\left(g\right)\)

b, m_{dd sau pứ} = 15,3 + 164,25 = 179,55 (g)

c, \(C\%_{ddAlCl_3}=\dfrac{0,3.133,5.100\%}{179,55}=22,31\%\)

2)

a, \(m_{HCl}=54,75.20\%=10,95\left(g\right)\Rightarrow n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)

PTHH: Al₂O₃  + 6HCl → 2AlCl₃ + 3H₂O

Mol:      0,05        0,3          0,1

\(m_{Al_2O_3}=0,05.102=5,1\left(g\right)\)

b, m_{dd sau pứ} = 5,1 + 54,75 = 59,85 (g) 

\(C\%_{ddAlCl_3}=\dfrac{0,1.133,5.100\%}{59,85}=22,31\%\)

a) PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

b) Ta có: \(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)=n_{MgCl_2}\)

\(\Rightarrow m_{MgCl_2}=0,15\cdot95=14,25\left(g\right)\)

c) Theo PTHH: \(\left\{{}\begin{matrix}n_{HCl\left(p.ứ\right)}=0,3\left(mol\right)\\n_{H_2}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{HCl\left(p.ứ\right)}=0,3\cdot36,5=10,95\left(g\right)\\m_{H_2}=0,15\cdot2=0,3\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{Mg}+m_{ddHCl}-m_{H_2}=53,3\left(g\right)\)

\(\Rightarrow m_{ddHCl}=50\left(g\right)\) \(\Rightarrow C\%_{HCl\left(p.ứ\right)}=\dfrac{10,95}{50}\cdot100\%=21,9\%\)  

Đáp án B

38,92 gam

Đáp án D

Gọi x = nHOOC-CH2-CH2-CH(NH2)-COOH; y = nCH3-CH(NH2)-COOH

m = 0,18.147 + 0,14.89 = 38,92

Chọn B