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a) nAl2O3= 5,1/102=0,5(mol)
PTHH: 4Al +3 O2 -to-> 2 Al2O3
nO2= 3/2. nAl2O3= 3/2 . 0,05= 0,075(mol)
=>mO2=0,075 x 32= 2,4(g)
b) 2 KMnO4 -to-> K2MnO4 + MnO2 + O2
nKMnO4= 2.nO2= 2. 0,075= 0,15(mol)
=> mKMnO4= 0,15 x 158= 23,7(g)
nAl2O3 = \(\dfrac{5,1}{102}\)=0,05
PTHH
4Al + 3O2 = 2Al2O3
theo phương trình 4 mol : 3 mol : 2 mol
theo đề bài 0,075mol : 0,05mol
mO2 = 0,075. 32= 2,4g
PTHH
2KMnO4 ➜ K2MnO4 + MnO2 + O2
theo phương trình 2mol : 1mol : 1mol : 1mol
theo đề bài 0,05 mol : 0,075 mol
mKMnO4= 0,05. 158= 7,9g
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)
\(n_{Al}=\dfrac{5,4}{27}=0,2mol\)
\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)
0,2 0,15 ( mol )
\(V_{O_2}=0,15.22,4=3,36l\)
\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)
0,3 0,15 ( mol )
\(m_{KMnO_4}=0,3.158=47,4g\)
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\
pthh:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
0,2 0,15
\(V_{O_2}=0,15.22,4=3,36l\\
PTHH:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
0,3 0,15
\(m_{KMnO_4}=158.0,3=47,4g\)
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 )
\(n_{KClO_3}=\dfrac{0,3}{75\%}=0,4mol\)
\(m_{KClO_3}=0,4.122,5=49g\)
Bạn tách ra từng câu nhé!
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 )
\(m_{KClO_3}=n_{KClO_3}.M_{KClO_3}=0,5.122,5=61,25g\)
Bài 3:
Ta có: \(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)
PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
_____0,2____0,6____0,4 (mol)
\(\Rightarrow V_{H_2}=0,6.22,4=13,44\left(l\right)\)
\(m_{Fe}=0,4.56=22,4\left(g\right)\)
Bài 4:
a, \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
\(n_{O_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)
PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
Xét tỉ lệ: \(\dfrac{0,2}{4}< \dfrac{0,35}{5}\), ta được O2 dư.
Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{5}{4}n_P=0,25\left(mol\right)\Rightarrow n_{O_2\left(dư\right)}=0,35-0,25=0,1\left(mol\right)\)
\(\Rightarrow m_{O_2\left(dư\right)}=0,1.32=3,2\left(g\right)\)
b, Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,1\left(mol\right)\Rightarrow m_{P_2O_5}=0,1.142=14,2\left(g\right)\)
Lần sau bạn nên chia nhỏ câu hỏi ra nhé.
Bài 1:
a, \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
Theo PT: \(n_{O_2}=\dfrac{2}{3}n_{Fe}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b, \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
\(n_{KMnO_4}=2n_{O_2}=0,4\left(mol\right)\Rightarrow m_{KMnO_4}=0,4.158=63,2\left(g\right)\)
Bài 2:
a, \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
_____0,1___________0,1_____0,15 (mol)
\(m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\)
\(V_{H_2}=0,15.22,4=3,36\left(l\right)\)
b, PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Theo PT: \(n_{CuO}=n_{H_2}=0,15\left(mol\right)\Rightarrow m_{CuO}=0,15.80=12\left(g\right)\)
a, PT: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
Ta có: \(n_{Al}=\dfrac{10,2}{27}=\dfrac{17}{45}\left(mol\right)\)
b, Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}=\dfrac{17}{60}\left(mol\right)\)
\(\Rightarrow V_{O_2}=\dfrac{17}{60}.22,4\approx6,347\left(l\right)\)
c, Theo PT: \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=\dfrac{17}{90}\left(mol\right)\)
\(\Rightarrow m_{Al_2O_3}=\dfrac{17}{90}.102\approx19,267\left(g\right)\)
d, PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
Theo PT: \(n_{KMnO_4}=2n_{O_2}=\dfrac{17}{30}\left(mol\right)\)
\(\Rightarrow m_{KMnO_3}=\dfrac{17}{30}.158\approx89,53\left(g\right)\)
a, \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
b, \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
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, \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
Theo PT: \(n_{KMnO_4}=2n_{O_2}=0,3\left(mol\right)\Rightarrow m_{KMnO_4}=0,3.158=47,4\left(g\right)\)
\(n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ PTHH:4Al+3O_2-^{t^o}>2Al_2O_3\)
tỉ lệ 4 : 3 ; 2
n(mol) 0,2----->0,15---->0,1
\(V_{O_2\left(dktc\right)}=n\cdot22,4=0,15\cdot22,4=3,36\left(l\right)\\ PTHH:2KMnO_4-^{t^o}>K_2MnO_4+MnO_2+O_2\)
tỉ lệ 2 : 1 ; 1 ; 1
n(mol) 0,3<------------------------------------------0,15
\(m_{KMnO_4}=n\cdot M=0,3\cdot\left(39+55+16\cdot4\right)=47,4\left(g\right)\)
a, PTHH ( I ) : \(4Al+3O_2\rightarrow2Al_2O_3\)
\(n_{Al_2O_3}=\frac{m_{Al_2O_3}}{M_{Al_2O_3}}=\frac{5,2}{27.2+16.3}=\frac{5,2}{102}\approx0,05\left(mol\right)\)
- Theo PTHH ( I ) : \(n_{O_2}=\frac{3}{2}.n_{Al_2O_3}=\frac{3}{2}.0,05=0,075\left(mol\right)\)
-> \(m_{O_2}=n.M=0,075.32=2,4\left(g\right)\)
b, PTHH ( II ) : \(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)
- Theo PTHH ( II ) : \(n_{KMnO_4}=2n_{O_2}=2.0,075=0,15\left(mol\right)\)
-> \(m_{KMnO_4}=n.M=0,15.158=23,7\left(g\right)\)
4Al+3O2--->2Al2O3
0,1--0,075----0,05 mol
nAl2O3=5,2 \102=0,05 mol
=>mO2=0,075.32=2,4 g
K2MnO4
0,15--------------------0,075 mol
=>mKMnO4=0,15.158 =23,7 g