2, a, \(a+\dfrac{1}{a}\ge2\)

\(\Leftrightarrow\dfrac{a^2+1}{a}\ge2\)

\(\Rightarrow a^2-2a+1\ge0\left(a>0\right)\)

\(\Leftrightarrow\left(a-1\right)^2\ge0\)( là đt đúng vs mọi a)

vậy...................

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

2Al + 3H₂SO₄ \(\rightarrow\) Al₂(SO₄)₃ + 3H₂

de: 0,2 \(\leftarrow\) 0,3 (mol)

vì Ag ko tac dung đc vs H₂SO₄ và Al₂(SO₄)₃ tan nên chất rắn k tan sau pu là Ag \(\Rightarrow m_{Ag}=4,6g\)

\(m_{Al}=0,2.27=5,4g\)

\(\%m_{Al}=\dfrac{5,4}{5,4+4,6}.100\%=54\%\)

\(\%m_{Ag}=100-54=46\%\)

b, \(CaCO_3\underrightarrow{t^o}CaO+CO_2\) (1)

CO₂ + 2NaOH \(\rightarrow\) Na₂CO₃ + H₂O (2)

Na₂CO₃+H₂SO₄\(\rightarrow\)Na₂SO₄+ CO₂+H₂O (3)

Na₂SO₄ + BaCl₂ \(\rightarrow\) 2NaCl + BaSO₄ (4)

2NaCl + 2H₂O \(\rightarrow\) 2NaOH + Cl₂ + H₂ (5)

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

Mg + 2HCl \(\rightarrow\) MgCl₂ + H₂

de: 0,1 \(\leftarrow\) 0,2 \(\leftarrow\) 0,1 \(\leftarrow\) 0,1 (mol)

\(m_{Mg}=0,1.24=2,4g\)

\(m_{MgO}=4,4-2,4=2g\)

\(n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\)

MgO + 2HCl \(\rightarrow\) MgCl₂ + H₂O

de: 0,05 \(\rightarrow\) 0,1 \(\rightarrow\) 0,05 \(\rightarrow\) 0,05 (mol)

\(m_{HCl}=36,5.\left(0,2+0,1\right)=10,95g\)

b, \(m_{ddHCl}=\dfrac{10,95.100}{7,3}=150g\)

\(m_{ct}=95.\left(0,1+0,05\right)=14,25g\)

\(m_{ddspu}=4,4+10,95-0,2=15,15g\)

c, \(C\%=\dfrac{14,25}{15,15}.100\%\approx94,06\%\)

bài 2 ở cuối cùng:

đổi 600ml = 0,6l

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

2Al + 6HCl \(\rightarrow\) 2AlCl₃ + 3H₂

de: 0,4 \(\rightarrow\) 1,2 \(\rightarrow\) 0,4 \(\rightarrow\) 0,6 (mol)

a, \(V_{H_2}=0,6.22,4=13,44l\)

b, \(m_{AlCl_3}=0,4.133,5=53,4g\)

c, \(C_{M_{HCl}}=\dfrac{1,2}{0,6}=2M\)

FeO + H₂ \(\rightarrow\) Fe + H₂O

de: 0,6 \(\leftarrow\) 0,6 (mol)

d, \(m_{FeO}=0,6.72=43,2g\)

bạn cứ phân tích ra như thế này:
gọi số đó là :ab
ab = 8 x (a+b)
10a + b= 8a + 8b
2 x a= 7 x b
vậy ab = 72

2, gọi x la so mol cua Al

y la so mol cua Mg

2Al + 3H₂SO₄\(\rightarrow\) Al₂(SO₄)₃ + 3H₂

de: x \(\rightarrow\) \(\dfrac{3x}{2}\) (mol)

Mg + H₂SO₄ \(\rightarrow\) MgSO₄ + H₂

de: y \(\rightarrow\) y (mol)

\(n_{H_2}=\dfrac{5,04}{22,4}=0,225\left(mol\right)\)

Ta co: 27x + 24y = 4,5

\(\dfrac{3x}{2}+y=0,225\)

\(\Rightarrow x=0,1\) ; y = 0,075

\(m_{Al}=0,1.27=2,7g\)

\(\%m_{Al}=\dfrac{2,7}{4,5}.100\%=60\%\)

\(\%m_{Mg}=100\%-60\%=40\%\)

goi x la so mol cua Al

y la so mol cua Fe

\(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

2Al + 6HCl \(\rightarrow\) 2AlCl₃ + 3H₂

de: x \(\rightarrow\) 1,5x

2Fe + 6HCl \(\rightarrow\) 2AlCl₃ + 3H₂

de: y \(\rightarrow\) 1,5y

Ta co: 27x + 56y = 16,6

1,5x + 1,5y = 0,5

\(\Rightarrow\left\{{}\begin{matrix}x\approx0,071\\y\approx0,262\end{matrix}\right.\)

a, \(m_{Al}=27.0,071\approx1,92g\)

\(m_{Fe}=16,6-1,92=14,68g\)

câu b bn tự lm nha

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Al₂O₃ + 6HCl \(\rightarrow\) 2AlCl₃ + 3H₂O

Cu + HCl ( Khong pu)

\(\Rightarrow\) kim loai A la Cu

\(\Rightarrow n_{CuO}=\dfrac{2,75}{80}\approx0,03\left(mol\right)\)

2Cu + O₂ \(\underrightarrow{t^0}\) 2CuO

de: 0,03 \(\leftarrow\) 0,03

\(m_{Cu}=1,92g\)

2Al + 6HCl \(\rightarrow\) 2AlCl₃ + 3H₂

de: 0,1 \(\leftarrow\) 0,15

\(m_{Al}=2,7g\)

\(m_{Al_2O_3}=10-2,7-1,92=5,38g\)

\(\%m_{Cu}=19,2\%\)

\(\%m_{Al}=27\text{%}\)

\(\%m_{Al_2O_3}=100-27-19,2=53,8\%\)

\(m_{H_2SO_4}=\dfrac{11.20}{100}2,2g\)

\(\Rightarrow n_{H_2SO_4}=\dfrac{2,2}{98}\approx0,02\left(mol\right)\)

\(m_{BaCl_2}=20,8g\Rightarrow n_{BaCl_2}=0,1\left(mol\right)\)

H₂SO₄ + BaCl₂ \(\rightarrow\) BaSO₄ + 2HCl

de: 0,02 0,1

pu: 0,02 0,02 0,02 0,04

spu: 0 0,08 0,02 0,04

\(m_{ddspu}=11+400-m_{BaSO_4}=406,34g\)

\(m_{ct}=m_{BaSO_4}+m_{BaCl_2dư}=21,3g\)

\(C\%=\dfrac{m_{ct}}{m_{ddspu}}.100\%\approx5,24\%\)