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\(n_{O_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\\ n_{KMnO_4}=\dfrac{15,8}{158}=0,1\left(mol\right)\\ 2KMnO_4\underrightarrow{^{to}}K_2MnO_4+MnO_2+O_2\\ n_{O_2\left(LT\right)}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ H=\dfrac{0,03}{0,05}.100=60\%\)
\(n_{KClO_3}=\dfrac{24,5}{122,5}=0,2\left(mol\right)\\ 2KClO_3\underrightarrow{^{to}}2KCl+3O_2\\ n_{O_2\left(LT\right)}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\\ Vì:H=90\%\Rightarrow n_{O_2\left(TT\right)}=90\%.0,3=0,27\left(mol\right)\\ V_{O_2\left(đktc,thực.tế\right)}=0,27.22,4=6,048\left(l\right)\)
\(n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\\ n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\\ 4P+5O_2\underrightarrow{^{to}}2P_2O_5\\ Vì:\dfrac{0,6}{5}>\dfrac{0,2}{1}\\ \Rightarrow O_2dư\\ n_{P_2O_5\left(LT\right)}=\dfrac{2}{4}.0,2=0,1\left(mol\right)\\ n_{P_2O_5\left(TT\right)}=0,1.75\%=0,075\left(mol\right)\\ m_{P_2O_5\left(TT\right)}=142.0,075=10,65\left(g\right)\)
\(n_P=\dfrac{6,2}{31}=0,2\left(mol\right);n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
PTHH: 4P + 5O2 ---to→ 2P2O5
Mol: 0,2 0,1
Ta có:\(\dfrac{0,2}{4}< \dfrac{0,6}{5}\) ⇒ P hết, O2 dư
\(m_{P_2O_5\left(lt\right)}=0,1.142=14,2\left(g\right)\Rightarrow m_{P_2O_5\left(tt\right)}=\dfrac{14,2}{75}.100=18,94\left(g\right)\)
\(Đặt:n_{KMnO_4\left(LT\right)}=a\left(mol\right)\\ 2KMnO_4\underrightarrow{^{to}}K_2MnO_4+MnO_2+O_2\\ Vì:m_{rắn}=29,04\\ \Leftrightarrow\left(31,6-158a\right)+197.0,5a+87.0,5a=29,04\\ \Leftrightarrow a=0,16\\ \Rightarrow H=\dfrac{0,16.158}{31,6}.100=80\%\)
\(Đặt:n_{KMnO_4\left(LT\right)}=a\left(mol\right)\\ 2KMnO_4\underrightarrow{^{to}}K_2MnO_4+MnO_2+O_2\\ n_{KMnO_4\left(bđ\right)}=\dfrac{31,6}{158}=0,2\left(mol\right)\\ n_{KMnO_4\left(LT\right)}=0,2-a\left(mol\right)\\ n_{K_2MnO_4}=n_{MnO_2}=0,5a\left(mol\right)\\ m_{rắn}=29,04\\ \Leftrightarrow m_{KMnO_4\left(LT\right)}+m_{K_2MnO_4}+m_{MnO_2}=29,04\\ \Leftrightarrow\left(31,6-158a\right)+197a.0,5+87a.0,5=29,04\\ \Leftrightarrow a=0,16\)
\(\Rightarrow H=\dfrac{0,16}{0,2}.100=80\%\)
2KClO3-to>2KCl+3O2
0,06-----------------0,09 mol
n O2=2,016\22,4=0,09 mol
=>H =0,06.122,5\12,25 .100=60%
\(n_{O_2\left(TT\right)}=\dfrac{2,016}{22,4}=0,09\left(mol\right)\\ n_{KClO_3}=\dfrac{12,25}{122,5}=0,1\left(mol\right)\\ 2KClO_3\underrightarrow{^{to}}2KCl+3O_2\\ n_{O_2\left(LT\right)}=\dfrac{3}{2}.0,1=0,15\left(mol\right)\\ \Rightarrow H=\dfrac{0,09}{0,15}.100=60\%\)
\(n_{SO2}=\dfrac{6,4}{64}=0,1\left(mol\right)\)
Pt : \(S+O_2\rightarrow\left(t_o\right)SO_2|\)
1 1 1
0,1 0,1
\(n_{O2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
\(V_{O2\left(lt\right)}=0,1.22,4=2,24\left(l\right)\)
⇒ \(V_{O2\left(tt\right)}=\dfrac{2,24.100}{80}=2,8\left(l\right)\)
Chúc bạn học tốt
\(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\\ n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\\ 3Fe+2O_2\underrightarrow{^{to}}Fe_3O_4\\ Vì:\dfrac{0,6}{2}>\dfrac{0,3}{3}\Rightarrow O_2dư\\ n_{Fe_3O_4\left(LT\right)}=\dfrac{0,3}{3}=0,1\left(mol\right)\\ n_{Fe_3O_4\left(TT\right)}=\dfrac{18,56}{232}=0,08\left(mol\right)\\ H=\dfrac{0,08}{0,1}.100=80\%\)
mC2H5OH(bd) = 0,8.23 = 18,4 (g)
=> \(n_{C_2H_5OH\left(bd\right)}=\dfrac{18,4}{46}=0,4\left(mol\right)\)
=> \(n_{C_2H_5OH\left(pư\right)}=\dfrac{0,4.75}{100}=0,3\left(mol\right)\)
PTHH: C2H5OH --H2SO4,170oC--> C2H4 + H2O
0,3------------------------->0,3
=> VC2H4 = 0,3.22,4 = 6,72 (l)
2KMnO4-to>K2MnO4+MnO2+O2
1,2-------------------------------------0,6 mol
n O2=13,44\22,4=0,6 mol
H =75%
=>m KMnO4 tt= 1,2.158 .100\75=252,8g
\(n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
PTHH: 2KMnO4 ---to→ K2MnO4 + MnO2 + O2
Mol: 1,2 0,6
\(m_{KMnO_4\left(lt\right)}=1,2.158=189,6\left(g\right)\Rightarrow m_{KMnO_4\left(tt\right)}=\dfrac{189,6}{75}.100=252,8\left(g\right)\)