a/ PTHH

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

b/

Ta có: \(n_{Al_2O_3}=\dfrac{10.2}{102}=0.1\left(mol\right)\)

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

4                   2

x                    0.1

\(=>x=\dfrac{0.1\cdot4}{2}=0.2=n_{Al}\)  

\(=>m_{Al}=0.2\cdot27=5.4\left(g\right)\)

a) \(n_{Al2O3}=\frac{20,4}{102}=0,2\left(mol\right)\)

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

0,4_____0,3____0,2(mol)

\(n_{Al}=0,4.27=10,8\left(g\right)\)

\(V_{O2}=0,3.22,4=6,72\left(l\right)\)

b) \(n_{O_2}=0,3\left(mol\right)\)

\(n_{Fe}=\frac{11,2}{56}=0,2\left(mol\right)\)

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

Ban đầu: 0,2____0,3

Phản ứng: 0,2____2/15__1/15 (mol)

Dư 1/6

So sánh: \(\frac{0,2}{3}< \frac{0,3}{2}\left(\frac{1}{15}< 0,15\right)\)

O2 dư

\(m_{Fe3O4}=\frac{1}{15}.232=15,46\left(g\right)\)

Chúc bạn học tốt

a) nAl2O3=20,4102=0,2(mol)

pT 4Al+3O2→2Al2O3

0,4_____0,3____0,2(mol)

nAl=0,4.27=10,8(g)

=>VO2=0,3.22,4=6,72(l)

b) nO2=0,3(mol)

nFe=11,256=0,2(mol)

PT 3Fe+2O2→Fe3O4

Ban đầu: 0,2____0,3

Phản ứng: 0,2____2/15__1/15 (mol)

==>Dư 1/6

So sánh: 0,2\3<0,3\2(0,O6<0,15)

==.O2 dư

=>mFe3O4=115.232=15,46(g)

a. \(n_{Al2O3}=\frac{20,4}{102}=0,2\left(mol\right)\)

\(PTHH:4Al+3O_2\rightarrow2Al_2O_3\)

_________0,4___0,3 ______0,2

\(\rightarrow m_{Al}=0,4.27=10,8\left(g\right)\)

\(\rightarrow m_{O2}=0,3.32=9,6\left(g\right)\)

b. \(n_{Fe}=\frac{11,2}{56}=0,2\left(mol\right)\)

\(PTHH:4Fe+3O_2\rightarrow2Fe_2O_3\)

Trước ___0,2 ___0,3_______

Phản ứng__ 0,2 ___0,15 ____________

Sau_______ 0 ____0,15 ___ 0,1

Nên lượng O2 đủ và còn dư 0,15mol

\(\rightarrow m_{Fe2O3}=0,1.160=16\left(g\right)\)

a) \(n_{Cu}=\dfrac{12,8}{64}=0,2\left(mol\right)\)

PTHH: 2Cu + O₂ --t^o--> 2CuO

             0,2-->0,1------->0,2

=> V_O2 = 0,1.22,4 = 2,24 (l)

b) m_CuO = 0,2.80 = 16 (g)

sorry e lớp 4

nên ko trả lời đc

a, Ta có: \(n_{Fe_3O_4}=\dfrac{11,6}{232}=0,05\left(mol\right)\)

PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

___0,15__0,1____0,05 (mol)

\(\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)

\(V_{O_2}=0,1.22,4=2,24\left(l\right)\)

b, PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

________0,2________________________0,1 (mol)

\(\Rightarrow m_{KMnO_4}=0,2.158=31,6\left(g\right)\)

Bạn tham khảo nhé!

\(n_{Al}=\dfrac{3,24}{27}=0,12mol\)

a)\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\) \(\Rightarrow\) phản ứng hóa hợp.

b)0,12    0,09    0,06

   \(m_{Al_2O_3}=0,06\cdot102=6,12g\)

c)\(V_{O_2}=0,09\cdot22,4=2,016l\)

a, Ta có: \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

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

Theo PT: \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,05\left(mol\right)\Rightarrow m_{Al_2O_3}=0,05.102=5,1\left(g\right)\)

b, Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}=0,075\left(mol\right)\Rightarrow V_{O_2}=0,075.22,4=1,68\left(l\right)\)

c, Có lẽ đề cho 0,112 chứ không phải 0,1121 bạn nhỉ?

Ta có: \(n_{O_2}=\dfrac{0,112}{22,4}=0,005\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,1}{4}>\dfrac{0,005}{3}\), ta được Al dư.

Theo PT: \(n_{Al_2O_3}=\dfrac{2}{3}n_{O_2}=\dfrac{1}{300}\left(mol\right)\Rightarrow m_{Al_2O_3}=\dfrac{1}{300}.102=0,34\left(g\right)\)

a.b.\(n_{KMnO_4}=\dfrac{m_{KMnO_4}}{M_{KMnO_4}}=\dfrac{15,8}{158}=0,1mol\)

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

   0,1                                                      0,05   ( mol )

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

c.\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\) 

              0,05               0,025  ( mol )

\(m_{Fe_3O_4}=n_{Fe_3O_4}.M_{Fe_3O_4}=0,025.232=5,8g\)

2KMnO4-to>K2MnO4+MnO2+O2

0,1-------------------------------------0,05

3Fe+2O2-to>Fe3O4

0,075--0,05----0,025

n KMnO4=\(\dfrac{15,8}{158}=0,1mol\)

=>VO2=0,05.22,4=1,12l

=>m Fe=0,075.56=4,2g

=>m Fe3O4=0,025.232=5,8g

4Al+3O2-to>2Al2O3

0,2----0,15-------0,1

nAl=\(\dfrac{5,4}{27}\)=0,2 mol

m Al2O3=0,1.102=10,2g

=>VO2=0,15.22,4=3,36l

2KClO3-to>2KCl+3O2

0,1----------------------0,15

=>m KClO3=0,1.122,5=12,25g

Làm gộp cả phần a và b

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,15mol\\n_{Al_2O_3}=0,1mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=0,15\cdot22,4=3,36\left(l\right)\\m_{Al_2O_3}=0,1\cdot102=10,2\left(g\right)\end{matrix}\right.\)