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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)
\(m_{Al_2O_3}=0,4.102=40,8\left(g\right)\)
\(V_{O_2}=0,6.22,4=13,44\left(l\right)\)
C+O2-to>CO2
4P+5O2-to>2P2O5
2H2+O2-to>2H2O
4Al+3O2-to>2Al2O3
3Fe+2O2-to>Fe3O4
2Mg+O2-to>2MgO
CH4+2O2-to>CO2+2H2O
C4H10+13\2O2to->4CO2+5H2O
\(C+O_2\underrightarrow{t^o}CO_2\\ 4P+5O_2\underrightarrow{t^o}2P_2O_5\\ 2H_2+O_2\underrightarrow{t^o}2H_2O\\ 4Al+3O_2\underrightarrow{t^o}2Al_2O_3\\ 3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\\ 2Mg+O_2\underrightarrow{t^o}2MgO\\ CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\\ 2C_4H_{10}+13O_2\underrightarrow{t^o}8CO_2+20H_2O\)
a)PTHH: \(2KClO_3\xrightarrow[MnO_2]{t^o}2KCl+3O_2\uparrow\)
b) Ta có: \(n_{KClO_3}=\dfrac{49}{122,5}=0,4\left(mol\right)\) \(\Rightarrow n_{O_2}=0,6\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,6\cdot22,4=13,44\left(l\right)\)
c) PTHH: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
Theo PTHH: \(n_P=\dfrac{4}{5}n_{O_2}=0,48\left(mol\right)\)
\(\Rightarrow m_P=0,48\cdot31=14,88\left(g\right)\)
4P+5O2-to>2P2O5
0,04----0,05----0,02
n P=0,04 mol
=>m P2O5=0,02.142=2,84g
=>VO2=0,05.22,4=1,12l
c)
2KMnO4-to>K2MnO4+MnO2+O2
0,1----------------------------------------0,05
H=10%
m KMnO4=0,1.158.110%=17,28g
\(n_P=\dfrac{1,24}{31}=0,04\left(mol\right)\\ pthh:4P+5O_2\underrightarrow{T^O}2P_2O_5\)
0,04 0,05 0,02
=> \(\left\{{}\begin{matrix}m_{P_2O_5}=0,02.142=2,84\left(g\right)\\V_{O_2}=0,05.22,4=1,12\left(l\right)\end{matrix}\right.\)
\(pthh:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
0,1 0,05
=> \(m_{KMnO_4}=0,1.158=15,8\left(g\right)\)
\(m_{KMnO_4\left(d\text{ùng}\right)}=15,8.110\%=17,38\left(g\right)\)
a) \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
PTHH: 4P + 5O2 --to--> 2P2O5
0,2-->0,25
=> VO2 = 0,25.22,4 = 5,6 (l)
=> Vkk = 5,6.5 = 28 (l)
b)
PTHH: 2KMnO4 --to--> K2MnO4 + MnO2 + O2
0,5<-----------------------------0,25
=> \(m_{KMnO_4}=0,5.158=79\left(g\right)\)
1) \(3O_2+4Al\underrightarrow{t^o}2Al_2O_3\)
2) \(O_2+4K\underrightarrow{t^o}2K_2O\)
3) \(5O_2+4P\underrightarrow{t^o}2P_2O_5\)
4) \(O_2+2H_2\underrightarrow{t^o}2H_2O\)
5) \(\dfrac{7}{2}O_2+C_2H_6\underrightarrow{t^o}2CO_2+3H_2O\)
\(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)
PTHH : 4Na + O2 -> 2Na2O
0,2---- 0,05 ---- 0,1 (mol)
\(V_{O_2}=0,05.22,4=1,12\left(l\right)\)
\(m_{Na_2O}=0,1.62=6,2\left(g\right)\)