Bài 1: n_Al2O3 = 30,6 / 102 = 0,3 (mol)

=> n_Al = 2n_Al2O3 = 0,6 mol

=> m_Al = 0,6 x 27 = 16,2 gam

=> m_O2 = 30,6 - 16,2 = 14,4 gam

Bài 2: n_Na = 2,3 / 23 = 0,1 (mol)

=> n_Na2SO4 = 0,05 mol

=> m_Na2SO4 = 0,05 x 142 = 7,1 gam

a/ %m_C = \(\frac{12}{12+16.2}.100\%=27,27\%\)

b/ %m_Al = \(\frac{27.2}{27.2+16.3}.100\%=52,94\%\)

=> %m_O = 100% - 52,94% = 47,06%

a/ %m_C = \(\dfrac{12}{12+16.2}\).100%=27,27%

b/ %m_Al = \(\dfrac{27.2}{27.2+16.3}\).100%=52,94%

%mO = 100% - 52,94% = 47,06%

Bài 1:

Fe + 2NaCl2 -> 2Na +FeCl2

FeCl2+ 2HCl -> FeCl3 +H2

2FeCl3 + 3Ca(OH)2 -> 2Fe(OH)3 +3CaCl2

Fe(OH)3 -> Fe2O3 +H2O

Fe2O3 ->2Fe + 3O2

Bài 1:

(1) Fe + 2HCl --> FeCl₂ + H₂

(2) 2FeCl₂ + Cl₂ --> 2FeCl₃

(3) \(FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3\downarrow+3NaCl\)

(4) \(2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H_2O\)

(5) \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)

Bài 2:

\(n_{Al_2O_3}=\dfrac{30,6}{102}=0,3\left(mol\right)\)

PTHH: Al₂O₃ + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂O

______0,3---------------------->0,3

=> \(m_{Al_2\left(SO_4\right)_3}=0,3.342=102,6\left(g\right)\)

Bài 3: 

2Al₂O₃ --npnc--> 4Al + 3O₂

=> 1 mol Al₂O₃ tạo ra 2 mol Al

=> 102g Al₂O₃ tạo ra 54g Al

=> 3,778g Al₂O₃ tạo ra 2g Al

=> 3,778 tấn Al₂O₃ tạo ra 2 tấn Al

=> m_{Al2O3(thực tế)} \(=\dfrac{3,778.100}{90}=4,198\left(tấn\right)\)

=> \(m_{boxit}=\dfrac{4,198.100}{40}=10,495\left(tấn\right)\)

bài 2 

n_Al2O3= \(\dfrac{30,6}{102}\)=0,3(mol)

Al₂O₃ + 3H₂SO₄ -> Al₂(SO₄)₃ + 3H₂O 

0,3                                 0,3

m_Al2(SO4)3  = 0,3*342=102,6 (g)

Trong $CO_2$ : $\%O = \dfrac{16.2}{44}.100\% = 72,73\%$

Trong $Al_2O_3$ : $\%O = \dfrac{16.3}{102}.100\% = 47,06\%$

Suy ra:  $\%O : CO_2 > Al_2O_3$

Câu 1:

a) Al₂O₃:

Phần trăm Al trong Al₂O₃:   \(\%Al=\dfrac{27.2}{27.2+16.3}.100=52,94\%\)

Phần trăm O trong Al₂O₃:   \(\%O=100-52,94=47,06\%\)

b) C₆H₁₂O:

Phần trăm C trong  C₆H₁₂O:  \(\%C=\dfrac{12.6}{12.6+12+16}.100=72\%\)

Phần trăm H trong  C₆H₁₂O:  \(\%H=\dfrac{1.12}{12.6+12+16}.100=12\%\)

Phần trăm O trong  C₆H₁₂O :  \(\%O=100-72-12=16\%\)

Câu 2: 

\(m_H=\dfrac{5,88.34}{100}\approx2\left(g\right)\)

\(m_S=\dfrac{94,12.34}{100}=32\left(g\right)\)

\(n_H=\dfrac{m}{M}=\dfrac{2}{1}=2\left(mol\right)\)

\(n_S=\dfrac{m}{M}=\dfrac{32}{32}=1\left(mol\right)\)

⇒ CTHH của hợp chất:  H₂S

a) \(M_{Ca\left(OH\right)_2}=40+\left(16+1\right).2=74\left(DvC\right)\)

\(\%Ca=\dfrac{40.1}{74}.100\%=54\%\)

\(\%O=\dfrac{16.2}{74}.100\%=43\%\)

\(\%H=100\%-54\%-43\%=3\%\)

a) \(\left\{{}\begin{matrix}\%m_{Ca}=\dfrac{40.1}{74}.100\%=54,054\%\\\%m_O=\dfrac{16.2}{74}.100\%=43,243\%\\\%m_H=\dfrac{2.1}{74}.100\%=2,703\%\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}\%m_{Ba}=\dfrac{137.1}{208}.100\%=65,865\%\\\%Cl=\dfrac{35,5.2}{208}.100\%=34,135\%\end{matrix}\right.\)

c) \(\left\{{}\begin{matrix}\%m_K=\dfrac{39.1}{56}.100\%=69,643\%\\\%m_O=\dfrac{16.1}{56}.100\%=28,571\%\\\%m_H=\dfrac{1.1}{56}.100\%=1,786\%\end{matrix}\right.\)

d) \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{27.2}{102}.100\%=52,94\%\\\%m_O=\dfrac{16.3}{102}.100\%=47,06\%\end{matrix}\right.\)

e) \(\left\{{}\begin{matrix}\%m_{Na}=\dfrac{23.2}{106}.100\%=43,396\%\\\%m_C=\dfrac{12}{106}.100\%=11,321\%\\\%m_O=\dfrac{16.3}{106}.100\%=45,283\%\end{matrix}\right.\)

g) \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{56.1}{72}.100\%=77,78\%\\\%m_O=\dfrac{16.1}{72}.100\%=22,22\%\end{matrix}\right.\)

h) \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{65.1}{161}.100\%=40,373\%\\\%m_S=\dfrac{32.1}{161}.100\%=19,876\%\\\%m_O=\dfrac{16.4}{161}.100\%=39,751\%\end{matrix}\right.\)

i) \(\left\{{}\begin{matrix}\%m_{Hg}=\dfrac{201.1}{217}.100\%=92,627\%\\\%m_O=\dfrac{16}{217}.100\%=7,373\%\end{matrix}\right.\)

k) \(\%m_{Na}=\dfrac{23.1}{85}.100\%=27,06\%;\%m_N=\dfrac{14.1}{85}.100\%=16,47\%\%;\%m_O=\dfrac{16.3}{85}.100\%=56,47\%\)

a, \(\%m_{Al}=\dfrac{54.100\%}{102}=53\%\)

\(\%m_O=100\%-53\%=47\%\)

b, \(\%m_{Ca}=\dfrac{40.100\%}{232}=17,24\%\)

\(\%m_H=\dfrac{2.100\%}{232}=0,86\%\)

\(\%m_P=\dfrac{62.100\%}{232}=26,72\%\)

\(\Rightarrow\%m_O=100\%-17,24\%-0,86\%-26,72\%=55,18\%\)

\(n_{Cr}=2n_{Cr_2O_3}=2\left(mol\right)\)

\(\Rightarrow m_{Cr}=104\left(g\right)\)

\(n_{Fe}=3n_{Fe_3O_4}=3\left(mol\right)\)

\(\Rightarrow m_{Fe}=168\left(g\right)\)

\(n_{Al}=2n_{Al_2O_3}=2\left(mol\right)\)

\(\Rightarrow m_{Al}=54\left(g\right)\)