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3Fe+2O2-to>Fe3O4
0,225--0,15
n Fe=\(\dfrac{12,6}{56}\)=0,225 mol
VO2=0,15.22,4=3,36l
2KClO3-to>2KCl+3O2
0,1---------------------0,15
=>m KClO3=0,1.122,5=12,25g
\(a,3Fe+2O_2\rightarrow Fe_3O_4\)
\(b,\)
Ta có : \(n_{Fe}=\dfrac{m}{M}=\dfrac{126}{56}=2,25\left(mol\right)\)
\(\Rightarrow n_{O_2}=\dfrac{2}{3}n_{Fe}=\dfrac{2}{3}.2,25=1,5\left(mol\right)\)
\(\Rightarrow VO_2=33,6\left(l\right)\)
\(c,\)
\(PTHH:2KClO_3\rightarrow2KCl+3O_2\)
Theo \(PTHH:n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{2}{3}.1,5=1\left(mol\right)\)
\(\Rightarrow m_{KClO_3}=n.M=1,122,5=122,5\left(g\right)\)
\(n_{CO_2}=\dfrac{4.4}{44}=0.1\left(mol\right)\)
\(CH_4+2O_2\underrightarrow{^{^{t^0}}}CO_2+2H_2O\)
\(0.1.......0.2........0.1..........0.2\)
\(m_{CH_4}=0.1\cdot16=1.6\left(g\right)\)
\(V_{H_2O}=0.2\cdot22.4=4.48\left(l\right)\)
\(V_{kk}=5V_{O_2}=5\cdot0.2\cdot22.4=22.4\left(l\right)\)
\(n_P=\dfrac{7,44}{31}=0,24mol\)
\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
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, \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
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: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=0,15\left(mol\right)\Rightarrow m_{KClO_3}=0,15.122,5=18,375\left(g\right)\)
\(n_{Fe}=\dfrac{25,2}{56}=0,45\left(mol\right)\\ a,PTHH:3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\\ b,n_{O_2}=\dfrac{2}{3}.0,45=0,3\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ c,2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\\ n_{KClO_3}=\dfrac{2}{3}.0,3=0,2\left(mol\right)\\ \Rightarrow m_{KClO_3}=122,5.0,2=24,5\left(g\right)\)
a) PTHH: \(Zn+\dfrac{1}{2}O_2\xrightarrow[]{t^o}ZnO\)
b) Ta có: \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\) \(\Rightarrow n_{O_2}=0,15\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,15\cdot22,4=3,36\left(l\right)\)
c) PTHH: \(KClO_3\xrightarrow[MnO_2]{t^o}KCl+\dfrac{3}{2}O_2\uparrow\)
Theo PTHH: \(n_{KClO_3}=0,1\left(mol\right)\) \(\Rightarrow m_{KClO_3}=0,1\cdot122,5=12,25\left(g\right)\)
\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25mol\)
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)
0,25 0,5 0,25 ( mol )
\(m_{CH_4}=0,25.16=4g\)
\(V_{O_2}=0,5.22,4=11,2l\)
TK