\(a,PTHH:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\\ n_{Al}=\dfrac{m}{M}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ Theo.PTHH:n_{Al_2O_3}=\dfrac{1}{2}.n_{Al}=\dfrac{1}{2}.0,4=0,2\left(mol\right)\\ m_{Al_2O_3}=n.M=0,2.102=20,4\left(g\right)\)

\(b,n_{O_2}=\dfrac{V_{\left(đktc\right)}}{22,4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ Lập.tỉ.lệ:\dfrac{n_{Al}}{4}>\dfrac{n_{O_2}}{3}\Rightarrow Al.dư\\ Theo.PTHH:n_{Al\left(pư\right)}=\dfrac{4}{3}.n_{O_2}=\dfrac{4}{3}.0,2\left(mol\right)\\ n_{Al\left(dư\right)}=n_{Al\left(bđ\right)}-n_{Al\left(pư\right)}=0,4-0,2=0,2\left(mol\right)\\ Theo.PTHH:n_{Al_2O_3}=\dfrac{1}{2}.n_{Al}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\\ m_{Al_2O_3}=n.M=0,1=102=10,2\left(g\right)\)

Bài 2:

\(n_{Al}=\dfrac{16,2}{27}=0,6\left(mol\right)\\ n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\\ a,4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ Vì:\dfrac{0,6}{4}< \dfrac{0,6}{3}\Rightarrow O_2dư\\ n_{O_2\left(dư\right)}=0,6-\dfrac{3}{4}.0,6=0,15\left(mol\right)\\ \Rightarrow m_{O_2\left(dư\right)}=0,15.32=4,8\left(g\right)\\ c,n_{Al_2O_3}=\dfrac{2}{4}.n_{Al}=\dfrac{2}{4}.0,6=0,3\left(mol\right)\\ \Rightarrow m_{Al_2O_3}=102.0,3=30,6\left(g\right)\)

Bài 1.

\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1mol\)

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

0,1             0,6          0,2

\(m_{HCl}=0,6\cdot36,5=21,9g\)

\(m_{FeCl_3}=0,2\cdot162,5=32,5g\)

\(a,n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

b, LTL: \(\dfrac{0,4}{4}>\dfrac{0,6}{3}\) => O₂ dư

Theo pthh: \(\left\{{}\begin{matrix}n_{O_2\left(pư\right)}=\dfrac{3}{4}n_{Al}=\dfrac{3}{4}.0,4=0,3\left(mol\right)\\n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=\dfrac{1}{2}.0,4=0,2\left(mol\right)\end{matrix}\right.\)

=> V_{O2 (dư)} = (0,6 - 0,3).22,4 = 6,72 (l)

c, m_Al2O3 = 0,2.102 = 20,4 (g)

\(n_{Al}=\dfrac{10,8}{17}=0,4\left(mol\right)\\ n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\\ pthh:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\) 
\(LTL:\dfrac{0,4}{4}< \dfrac{0,6}{3}\) 
=> O2 dư P hết 
\(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,2\left(mol\right)\\ m_{Al_2O_3}=0,2.102=20,4g\)

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

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

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

Xét: \(\dfrac{0,4}{4}\) < \(\dfrac{0,6}{3}\)                  ( mol )

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

Chất dư là O₂

\(m_{O_2\left(dư\right)}=\left(0,6-0,3\right).32=9,6g\)

\(m_{Al_2O_3}=0,2.102=20,4g\)

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

b)

\(n_{Al} = \dfrac{21,6}{27} = 0,8(mol)\)

Theo PTHH :

\(n_{Al_2O_3} = \dfrac{1}{2}n_{Al} = 0,4(mol)\\ \Rightarrow m_{Al_2O_3} = 0,4.102 = 40,8(gam)\)

c)

\(n_{O_2} = \dfrac{3}{4}n_{Al} = 0,6(mol)\\ \Rightarrow V_{O_2} = 0,6.22,4 = 13,44(lít)\\ \Rightarrow V_{không\ khí} = 5V_{O_2} = 13,44.5 = 67,2(lít)\)

\(n_{Al}=\dfrac{21.6}{27}=0.8\left(mol\right)\)

\(4Al+3O_2\underrightarrow{t^0}2Al_2O_3\)

\(0.8......0.6........0.4\)

\(m_{Al_2O_3}=0.4\cdot102=40.8\left(g\right)\)

\(V_{kk}=5V_{O_2}=5\cdot0.6\cdot22.4=67.2\left(l\right)\)

Đốt cháy 10,8g nhôm trong bình chứa oxi.

a,viết phương trình hoá học của phản ứng. 

b,tính thể tích khí oxi cần dùng (ở đktc)

c,tính khối lượng KMnO4 cần dùng để điều chế lượng oxi nói trên

\(n_{KMnO_4}=\dfrac{18.96}{158}=0.12\left(mol\right)\)

\(2KMnO_4\underrightarrow{t^0}K_2MnO_4+MnO_2+O_2\)

\(0.12...........................................0.06\)

\(V_{O_2}=0.06\cdot22.4=1.344\left(l\right)\)

\(n_{Al}=\dfrac{5.4}{27}=0.2\left(mol\right)\)

\(4Al+3O_2\underrightarrow{t^0}2Al_2O_3\)

\(0.08.....0.06.......0.04\)

\(m_{Al\left(dư\right)}=\left(0.2-0.08\right)\cdot27=3.24\left(g\right)\)

\(m_{Al_2O_3}=0.04\cdot102=4.08\left(g\right)\)

a: \(4Al+3O_2\rightarrow2Al_2O_3\)

c: \(n_{Al}=\dfrac{2.4}{24}=0.1\left(mol\right)\)

\(\Leftrightarrow n_{Al_2O_3}=0.05\left(mol\right)\)

\(m_{Al_2O_3}=0.05\cdot96=1.92\left(g\right)\)

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

Tỉ lệ: n_Al : n_O2 = 4:3

b, Phần này bạn xem lại đề 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)\(V_{O_2}=0,09\cdot22,4=2,016l\)

Bài 1:

\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

\(4P+5O_2\rightarrow2P_2O_5\)

0,24.... 0,3 .... 0,12 (mol)

\(m_P=0,24.31=7,44\left(g\right)\)

\(m_{P_2O_5}=0,12.142=17,04\left(g\right)\)

Bài 2:

\(n_{Al}=\dfrac{21,6}{27}=0,8\left(mol\right)\)

\(4Al+3O_2\rightarrow2Al_2O_3\)

0,8 .... 0,6 ...... 0,4 (mol) 

\(m_{Al_2O_3}=0,4.102=40,8\left(g\right)\)

\(V_{O_2}=0,6.22,4=13,44\left(l\right)\)