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) 

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

nAl2O3= 10,2/102= 0,1(mol)

a) PTHH: Al2O3 + 6 HCl -> 2 AlCl3 + 3 H2O

0,1_______0,6_______0,2_________0,3(mol)

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

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

b) mddsau= mAl2O3 + mddHCl= 10,2+300=310,2(g)

c) mAlCl3= 133,5.0,2=26,7(g)

=>C%ddAlCl3= (26,7/310,2).100=8,607%

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

Đáp án B

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

=> m = 0,18.147 + 0,14.89 = 38,92 

a)

$2Al + 6HCl \to 2AlCl_3 + 3H_2$

$n_{Al} = \dfrac{10,8}{27} = 0,4(mol)$

Theo PTHH : $n_{HCl} = 3n_{Al} = 1,2(mol)$
$\Rightarrow m = \dfrac{1,2.36,5}{14,6\%} = 300(gam)$

b) $n_{H_2} = \dfrac{3}{2}n_{Al} = 0,6(mol)$

$M_xO_y + yH_2 \xrightarrow{t^o}xM + yH_2O$

Theo PTHH : $n_{oxit} = \dfrac{1}{y}.n_{H_2} = \dfrac{0,6}{y}(mol)$
$\Rightarrow \dfrac{0,6}{y}(Mx + 16y) = 34,8$

$\Rightarrow \dfrac{x}{y}.M = 42$

Với x = 3 ; y = 4 thì $M = 56(Fe)$

Vậy oxit là $Fe_3O_4$

\(n_{HCl}=0,1.0,3=0,03\left(mol\right)\)

=> \(n_{H_2O}=\dfrac{0,03}{2}=0,015\left(mol\right)\)

Theo ĐLBTKL: \(m_{oxit}+m_{HCl}=m_{muối}+m_{H_2O}\)

=> m_muối = 3,425 + 0,03.36,5 - 0,015.18 = 4,25(g) 

\(m_{ct}=\dfrac{20.98}{100}=19,6\left(g\right)\)

\(n_{H2SO4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)

Pt : \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O|\)

         1             1               1             1

        0,2          0,2           0,2

a) \(n_{CuO}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{CuO}=0,2.80=16\left(g\right)\)

b) \(n_{CuSO4}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{CuSO4}=0,2.160=32\left(g\right)\)

c) \(m_{ddspu}=16+98=114\left(g\right)\)

\(C_{CuSO4}=\dfrac{32.100}{114}=28,7\)⁰/₀

 Chúc bạn học tốt

\(n_{Fe_2O_3}=\dfrac{8}{160}=0,05\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂SO₄ --> Fe₂(SO₄)₃ + 3H₂O

______0,05------>0,15--------->0,05

=> m_H2SO4 = 0,15.98 = 14,7(g)

=> \(C\%\left(H_2SO_4\right)=\dfrac{14,7}{100}.100\%=14,7\%\)

\(C\%\left(Fe_2\left(SO_4\right)_3\right)=\dfrac{0,05.400}{8+100}.100\%=18,52\%\)

PTHH: Fe₂(SO₄)₃ + 6NaOH --> 2Fe(OH)₃\(\downarrow\) + 3Na₂SO₄

________0,05----------------------->0,1

=> mFe(OH)₃ = 0,1.107=10,7(g)