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https://hoc24.vn/cau-hoi/cho-324-g-al-tac-dung-voi-oxi-vua-du-th-duoc-al2o3-a-tinh-vo2-b-tinh-m-al2o3-c-trong-vkk-can-dung-biet-vo2-21-vkk-d-tinh-khoi-luong-kmno.7651142171785
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)
nFe = 46,4/56 = 29/35 (mol)
PTHH: 4Fe + 3O2 -> (t°) 2Fe2O3
Mol: 29/35 ---> 87/140 ---> 29/70
mFe2O3 = 29/70 . 160 = 464/7 (g)
Vkk = 87/140 . 5 . 22,4 = 69,6 (l)
\(n_{Al}=\dfrac{6,75}{27}=0,25mol\)
\(n_{O_2}=\dfrac{3,36}{22,4}=0,15mol\)
\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
0,25 0,15 0
0,2 0,15 0,1
0,05 0 0,1
\(m_{dư}=m_{Aldư}=0,05\cdot27=1,35g\)
\(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)
0,3 0,15
\(m_{KMnO_4}=0,3\cdot158=47,4g\)
nCu = 6,4/64 = 0,1 (mol)
PTHH: 2Cu + O2 -> (t°) 2CuO
Mol: 0,1 ---> 0,05 ---> 0,1
mCuO = 0,1 . 80 = 8 (g)
Vkk = 0,05 . 5 . 22,4 = 5,6 (l)
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.\(n_{C_4H_{10}}=\dfrac{4,48}{22,4}=0,2mol\)
\(2C_4H_{10}+13O_2\rightarrow\left(t^o\right)8CO_2+10H_2O\)
0,2 1,3 ( mol )
\(V_{O_2}=1,3.22,4=29,12l\)
\(V_{kk}=29,12.5=145,6l\)
b.\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)
2,6 1,3 ( mol )
\(m_{KMnO_4}=2,6.158=410,8g\)
\(n_{C_4H_{10}}=\dfrac{4,48}{22,4}=0,2\left(MOL\right)\\
pthh:2C_2H_{10}+13O_2\underrightarrow{t^o}8CO_2+10H_2O\)
0,2 1,3
=> \(V_{O_2}=1,3.22,4=29,12\left(l\right)\\
V_{kk}=29,12:20\%=145,6\left(l\right)\\
pthh:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
2,6 1,3
=> \(m_{KMnO_4}=2,6.158=410,8\left(g\right)\)
PT: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
Ta có: \(n_{Al}=\dfrac{3,24}{27}=0,12\left(mol\right)\)
a, \(n_{O_2}=\dfrac{3}{4}n_{Al}=0,09\left(mol\right)\) \(\Rightarrow V_{O_2}=0,09.22,4=2,016\left(l\right)\)
b, \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,06\left(mol\right)\) \(\Rightarrow m_{Al_2O_3}=0,06.102=6,12\left(g\right)\)
c, \(V_{kk}=\dfrac{2,016}{21\%}=9,6\left(l\right)\)
d, PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
Theo PT: \(n_{KMnO_4}=2n_{O_2}=0,18\left(mol\right)\)
\(\Rightarrow m_{KMnO_4}=0,18.158=28,44\left(g\right)\)