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Bài 3.
\(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{36}{56}=0,6428mol\)
\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)
0,6428 ----- 0,4285 ( mol )
\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)
0,857 0,4285 ( mol )
\(m_{KMnO_4}=n_{KMnO_4}.M_{KMnO_4}=0,857.158=135,406g\)
Bài 4.
a.\(n_{Al_2O_3}=\dfrac{m_{Al_2O_3}}{M_{Al_2O_3}}=\dfrac{51}{102}=0,5mol\)
\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)
1 0,75 0,5 ( mol )
\(m_{Al}=n_{Al}.M_{Al}=1.27=27g\)
\(V_{O_2}=n_{O_2}.22,4=0,75.22,4=16,8l\)
b.\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)
1,5 0,75 ( mol )
\(m_{KMnO_4}=n_{KMnO_4}.M_{KMnO_4}=1,5.158=237g\)
\(2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\)
0,5 0,75 ( mol )
\(m_{KClO_3}=n_{KClO_3}.M_{KClO_3}=0,5.122,5=61,25g\)
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 )
\(n_{KClO_3}=\dfrac{0,3}{75\%}=0,4mol\)
\(m_{KClO_3}=0,4.122,5=49g\)
a, \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
b, \(n_{Al_2O_3}=\dfrac{20,4}{102}=0,2\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{3}{2}n_{Al_2O_3}=0,3\left(mol\right)\Rightarrow V_{O_2}=0,3.22,4=6,72\left(l\right)\)
c, \(V_{kk}=\dfrac{V_{O_2}}{20\%}=33,6\left(l\right)\)
a, \(2Zn+O_2\underrightarrow{t^o}2ZnO\)
b, \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{Zn}=0,1\left(mol\right)\Rightarrow V_{O_2}=0,1.22,4=2,24\left(l\right)\)
\(\Rightarrow V_{kk}=5V_{O_2}=11,2\left(l\right)\)
c, \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
Theo PT: \(n_{KMnO_4}=2n_{O_2}=0,2\left(mol\right)\Rightarrow m_{KMnO_4}=0,2.158=31,6\left(g\right)\)
a) $n_{Al} = \dfrac{5,4}{27} = 0,2(mol)$
$4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3$
Theo PTHH : $n_{O_2} = \dfrac{3}{4}n_{Al} = 0,15(mol)$
$V_{O_2} = 0,15.22,4 = 3,36(lít)$
b) $2 KClO_3 \xrightarrow{t^o} 2KCl + 3O_2$
$n_{KClO_3} = \dfrac{2}{3}n_{O_2} = 0,1(mol)$
$m_{KClO_3} = 0,1.122,5 = 12,25(gam)$
\(n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ PTHH:4Al+3O_2-^{t^o}>2Al_2O_3\)
tỉ lệ: 4 : 3 : 2
n(mol) 0,2---->0,15---->0,1
\(V_{O_2\left(dktc\right)}=n\cdot22,4=0,15\cdot22,4=3,36\left(l\right)\\ PTHH:2KClO_3-^{t^o}>2KCl+3O_2\)
tỉ lệ: 2 : 2 : 3
n(mol) 0,1<-------------------------0,15
\(m_{KClO_3}=n\cdot M=0,1\cdot\left(39+35,5+16\cdot3\right)=12,25\left(g\right)\)
a, PT: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
b, Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}=0,15\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,15.22,4=3,36\left(l\right)\)
c, Theo PT: \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,1\left(mol\right)\)
\(\Rightarrow m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)
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: \(n_{Al_2O_3}=\dfrac{2}{3}n_{O_2}=\dfrac{1}{300}\left(mol\right)\Rightarrow m_{Al_2O_3}=\dfrac{1}{300}.102=0,34\left(g\right)\)
nAl=16,2/27= 0,6(mol)
a) PTHH: 4 Al +3 O2 -to-> 2 Al2O3
nO2= 3/4 . nAl=3/4 . 0,6= 0,45(mol)
=> V(O2,đktc)=0,45 x 22,4=10,08(l)
b) nAl2O3= nAl/2=0,6/2=0,3(mol)
=>mAl2O3=102. 0,3= 30,6(g)
c) 2KMnO4 -to-> K2MnO4 + MnO2 + O2
nKMnO4= 2.nO2=2. 0,45=0,9(mol)
=>mKMnO4= 158 x 0,9= 142,2(g)