n_Mg = 9,6/24 = 0,4 (mol)

2Mg + O2 ---t^o---> 2MgO

0,4____0,2_________0,4

V_O2(đktc) = 0,2.22,4 = 4,48(l)

m_MgO = 0,4.40 = 16(g)

a, PT: \(2Mg+O_2\underrightarrow{t^o}2MgO\)

b, Ta có: \(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)

Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{Mg}=0,15\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,15.22,4=3,36\left(l\right)\)

c, Theo PT: \(n_{MgO}=n_{Mg}=0,3\left(mol\right)\)

\(\Rightarrow m_{MgO}=0,3.40=12\left(g\right)\)

Bạn tham khảo nhé!

\(\sum\limits^{ }_{ }\)

a.b.\(n_{Mg}=\dfrac{m}{M}=\dfrac{6,4}{24}=\dfrac{4}{15}mol\)

\(2Mg+O_2\rightarrow\left(t^o\right)2MgO\)

4/15     2/15                       ( mol )

\(V_{O_2}=n.22,4=\dfrac{2}{15}.22,4=\dfrac{224}{75}l\)

c.\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)

       4/15                                                   2/15 ( mol )

\(m_{KMnO_4}=n.M=\dfrac{4}{15}.158=\dfrac{632}{15}g\)

nMg = 6,4 : 24= 0,26(mol) 
pthh : 2Mg+O2 -t--> 2MgO
       0,26 --> 0,13 (mol ) 
=> VO2(đktc) = 0,13.22,4=2,912(l)
pthh : 2KMnO4-t--> K2MnO4 + MnO2+ O2
                 0,26<------------------------------0,13(mol) 
 => mKMnO4 = 0,26.158= 41,08(g)

1. a. \(PTHH:2Mg+O_2\overset{t^o}{--->}2MgO\left(1\right)\)

b. Ta có: \(n_{Mg}=\dfrac{24}{24}=1\left(mol\right)\)

Theo PT₍₁₎: \(n_{O_2}=\dfrac{1}{2}.n_{Mg}=\dfrac{1}{2}.0,1=0,5\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,5.22,4=11,2\left(lít\right)\)

c. \(PTHH:2KClO_3\xrightarrow[t^o]{MnO_2}2KCl+3O_2\left(2\right)\)

Theo PT₍₂₎: \(n_{KClO_3}=\dfrac{2}{3}.n_{O_2}=\dfrac{2}{3}.0,5=\dfrac{1}{3}\left(mol\right)\)

\(\Rightarrow m_{KClO_3}=\dfrac{1}{3}.122,5=40,83\left(g\right)\)

2. \(PTHH:3Fe+2O_2\overset{t^o}{--->}Fe_3O_4\)

Ta có: \(n_{Fe_3O_4}=\dfrac{2,32}{232}=0,01\left(mol\right)\)

a. Theo PT: \(n_{Fe}=3.n_{Fe_3O_4}=0,01.3=0,03\left(mol\right)\)

\(\Rightarrow m_{Fe}=0,03.56=1,68\left(g\right)\)

b. Theo PT: \(n_{O_2}=2.n_{Fe_3O_4}=2.0,01=0,02\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,02.22,4=0,448\left(lít\right)\)

nAl = 2,7/27 = 0,1 (mol)

PTHH: 4Al + 3O2 -> (t°) 2Al2O3

Mol: 0,1 ---> 0,075 ---> 0,05

mAl2O3 = 0,05 . 102 = 5,1 (g)

VO2 = 0,075 . 22,4 = 1,68 (l)

Vkk = 1,68 . 5 = 8,4 (l)

\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{2,7}{27}=0,1mol\)

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

0,1     0,075           0,05          ( mol )

\(m_{Al_2O_3}=n_{Al_2O_3}.M_{Al_2O_3}=0,05.102=5,1g\)

\(V_{kk}=V_{O_2}.5=\left(0,075.22,4\right).5=8,4l\)

Giúp vs

a, \(2Mg+O_2\underrightarrow{^{t^o}}2MgO\)

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

\(n_{O_2}=\dfrac{1}{2}n_{MgO}=0,025\left(mol\right)\Rightarrow V_{O_2}=0,025.22,4=0,56\left(l\right)\)

b, Có lẽ đề cho oxi tác dụng với hidro chứ không phải oxit bạn nhỉ?

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

PT: \(2H_2+O_2\underrightarrow{^{t^o}}2H_2O\)

Xét tỉ lệ: \(\dfrac{0,15}{2}>\dfrac{0,025}{1}\), ta được H₂ dư.

THeo PT: \(n_{H_2O}=2n_{O_2}=0,05\left(mol\right)\Rightarrow m_{H_2O}=0,05.18=0,9\left(g\right)\)

a) 3Fe  +  2O₂      Fe₃O₄  

b) nFe = \(\dfrac{8,4}{56}\)= 0,15 mol 

nFe₃O₄ = \(\dfrac{11,6}{232}\) = 0,05 mol

Ta thấy \(\dfrac{nFe}{3}\)= \(\dfrac{nFe_3O_4}{1}\)=> Fe phản ứng hết 

<=> nO₂ cần dùng = \(\dfrac{2nFe}{3}\)= 0,1 mol 

<=> mO₂ cần dùng = 0,1.32 = 3,2 gam

c) Oxi chiếm thể tích bằng 1/5 thể tích không khí.

Mà V O₂ = 0,1.22,4 = 2,24 lít => V không khí = 2,24 . 5 = 11,2 lít

PTHH: \(2Mg+O_2\underrightarrow{t^o}2MgO\)

Bảo toàn khối lượng: \(m_{Mg}+m_{O_2}=m_{MgO}\)

\(\Rightarrow m_{O_2}=m_{MgO}-m_{Mg}=4-2,4=1,6\left(g\right)\)