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\(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\\ a,PTHH:4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\\ b,n_{O_2}=\dfrac{5}{4}.0,4=0,5\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,5.22,4=11,2\left(l\right)\\ c,n_{P_2O_5}=\dfrac{2}{4}.0,4=0,2\left(mol\right)\\ m_{P_2O_5}=142.0,2=28,4\left(g\right)\)
\(n_P=\dfrac{7,44}{31}=0,24mol\)
\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
0,24 0,3 0,12
\(V_{O_2}=0,3\cdot22,4=6,72l\)
\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
0,2 0,3
\(m_{KClO_3}=0,2\cdot122,5=24,5g\)
\(a,PTHH:2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\\ b,n_{KMnO_4}=\dfrac{15,8}{158}=0,1\left(mol\right)\\ n_{O_2}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,05.22,4=1,12\left(l\right)\\ c,4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\\ n_{P_2O_5}=\dfrac{2}{5}.n_{O_2}=\dfrac{2}{5}.0,05=0,02\left(mol\right)\\ m_{P_2O_5}=0,02.142=2,84\left(g\right)\)
a, \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
b, \(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\)
Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,2\left(mol\right)\Rightarrow m_{P_2O_5}=0,2.142=28,4\left(g\right)\)
c, \(n_{O_2}=\dfrac{5}{4}n_P=0,5\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,5.22,4=11,2\left(l\right)\)
d, \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
Theo PT: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{1}{3}\left(mol\right)\Rightarrow m_{KClO_3}=\dfrac{1}{3}.122,5=\dfrac{245}{6}\left(g\right)\)
a, \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
b, \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}=0,225\left(mol\right)\Rightarrow V_{O_2}=0,225.22,4=5,04\left(l\right)\)
c, \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
Theo PT: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=0,15\left(mol\right)\Rightarrow m_{KClO_3}=0,15.122,5=18,375\left(g\right)\)
nP= 0,2(mol)
a) PTHH: 4P + 5 O2 -to-> 2 P2O5
0,2_________0,25_____0,1(mol)
b) V(O2,đktc)=0,25 x 22,4= 5,6(l)
c) mP2O5=142 x 0,1=14,2(g)
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, \(V_{kk}=\dfrac{V_{O_2}}{20\%}=33,6\left(l\right)\)
2KMnO4-to->K2MnO4+MnO2+O2
0,6------------------------------------0,3 mol
nKMnO4=94,8\158=0,6 mol
=>VO2=0,3.22,4=6,72l
2.4P+5O2-to->2P2O5
0,8----1 -------------0,4 mmol
nP=24,8\31=0,8 mol
=>VO2=1.22,4=22,4l
=>mP2O5=0,4.142=56,8g