Bài này anh giúp rồi mà em. Em không hiểu chỗ nào nhỉ? 

\(n_{Al}=\dfrac{16,2}{27}=0,6mol\)

\(n_{O_2}=\dfrac{13,44}{22,4}=0,6mol\)

\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

0,6       0,6

Theo pt: \(n_{Al}=\dfrac{0,6}{4}=0,15< \dfrac{0,6}{3}=0,2=n_{O_2}\)

\(\Rightarrow O_2\) dư và dư \(0,6-\dfrac{0,6}{4}\cdot3=0,15mol\)

\(n_{Al_2O_3}=n_{Al}=0,15mol\)

\(m_{Al_2O_3}=0,15\cdot102=15,3g\)

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

\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)

 0,6     0,45           0,3                ( mol )

\(m_{Al}=0,6.27=16,2g\)

\(V_{O_2}=0,45.22,4=10,08l\)

\(V_{kk}=10,08.5=50,4l\)

b.\(2KClO_3\rightarrow\left(t^o,MnO_2\right)2KCl+3O_2\)

       0,3                                           0,45 ( mol )

\(m_{KClO_3}=0,3.122,5=36,75g\)

c.\(2KClO_3\rightarrow\left(t^o,MnO_2\right)2KCl+3O_2\)

       0,3                                           0,45  ( mol )

\(n_{KClO_3}=\dfrac{0,3}{75\%}=0,4mol\)

\(m_{KClO_3}=0,4.122,5=49g\)

Bài 2:

\(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\\ n_{Fe}=n_{H_2}=0,45\left(mol\right);n_{HCl}=2.0,45=0,9\left(mol\right)\\ a,m_{Fe}=0,45.56=25,2\left(g\right)\\ b,C_{MddHCl}=\dfrac{0,9}{0,15}=6\left(M\right)\)

\(m_{Mg}+m_{O_2}\rightarrow m_{MgO}\Leftrightarrow4,8g+m_{O_2}\rightarrow8\Leftrightarrow m_{O_2}=3,2g\)

MIK ĐANG CẦN GẤP GIÚP MIK VỚI

PTHH: \(2KMnO_4\xrightarrow[]{t^o}K_2MnO_4+MnO_2+O_2\)

\(n_{KMnO_4}=\dfrac{m_{KMnO_4}}{M_{KMnO_4}}=\dfrac{15,8}{158}=0,1\left(mol\right)\)

a. Theo PTHH: \(n_{O_2}=\dfrac{1}{2}n_{KMnO_4}=\dfrac{1}{2}0,1=0,05\left(mol\right)\)

\(\Rightarrow V_{O_2}=n_{O_2}.22,4=0,05.22,4=1,12\left(l\right)\)

b. PTHH: \(3Fe+2O_2\xrightarrow[]{t^o}Fe_3O_4\)

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

Ta có: \(\dfrac{1}{n_{O_2}}=\dfrac{1}{0,05}\)

\(\dfrac{1}{n_{Fe}}=\dfrac{1}{0,1}\)

\(\Rightarrow\dfrac{1}{n_{O_2}}>\dfrac{1}{n_{Fe}}\)

Vậy Fe dư

Theo PTHH: \(n_{Fe_3O_4}=\dfrac{0,1.1}{3}=\dfrac{1}{30}\left(mol\right)\)

\(\Rightarrow m_{Fe_3O_4}=n_{Fe_3O_4}.M_{Fe_3O_4}=\dfrac{1}{30}.232\approx7,73g\)

nMg = 5,76/24 = 0,24 (mol)

PTHH: 2Mg + O2 -> (t°) 2MgO

nMgO = 0,24 (mol)

mMgO = 0,24 . 40 = 9,6 (g)

nMg = 5,76 : 24 = 0,24 ( mol ) 
pthh : 2Mg+ O2 -t--> 2MgO
          0,24->0,12-->0,24 (mol) 
=> m = mMgO = 0,24 . 40 = 9,6 (g) 

\(n_{O_2}=\dfrac{0.896}{22.4}=0.04\left(mol\right)\)

\(3Fe+2O_2\underrightarrow{^{^{t^0}}}Fe_3O_4\)

\(0.06......0.04.......0.02\)

\(m_{Fe}=0.06\cdot56=3.36\left(g\right)\)

\(m_{Fe_2O_3}=0.02\cdot232=4.64\left(g\right)\)

3Fe+2O₂-to>Fe₃O₄

0,06----0,04-----0,02 mol

n _O2=\(\dfrac{0,896}{22,4}\)=0,04 mol

=>m _Fe=0,06.56=3,36g

=>m _Fe3O4=0,02.232=4,64g

a. -PTHH xảy ra: 3Fe+2O₂→Fe₃O₄.

-n_O2=\(\dfrac{V}{22,4}=\dfrac{0,896}{22,4}=0,04\)(mol) (bonus: ở trường mình là dùng 24,79 nhé:)

- Theo PTHH ta có:

3.n_Fe=2.n_O2=n_Fe3O4=0,04 (mol)

=>n_Fe=\(\dfrac{0,04}{3}=\dfrac{1}{75}\)(mol)

=>m_Fe=M.n=56.\(\dfrac{1}{75}\)=0,75(g).

b. n_Fe3O4=0,04 (mol)

=>m_Fe3O4=M.n=232.0,04=9,23(g)

nP = 3.1/31 = 0.1 (mol) 

4P + 5O2 -to-> 2P2O5 

0.1__0.125_____0.05

mP2O5 = 0.05*142 = 7.1 (g) 

VO2 = 0.125 * 22.4 = 2.8 (l)