\(n_{O_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\\ n_{KMnO_4}=\dfrac{15,8}{158}=0,1\left(mol\right)\\ 2KMnO_4\underrightarrow{^{to}}K_2MnO_4+MnO_2+O_2\\ n_{O_2\left(LT\right)}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ H=\dfrac{0,03}{0,05}.100=60\%\)

2KMnO4-to>K2MnO4+MnO2+O2

0,06----------------------------------0,03 mol

n O2=0,672\22,4=0,03 mol

=>H=0,06.158\15,8 .100=60%

\(n_{KClO_3}=\dfrac{24,5}{122,5}=0,2\left(mol\right)\\ 2KClO_3\underrightarrow{^{to}}2KCl+3O_2\\ n_{O_2\left(LT\right)}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\\ Vì:H=90\%\Rightarrow n_{O_2\left(TT\right)}=90\%.0,3=0,27\left(mol\right)\\ V_{O_2\left(đktc,thực.tế\right)}=0,27.22,4=6,048\left(l\right)\)

\(S+O_2\underrightarrow{^{to}}SO_2\\ n_{SO_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\\ n_{S\left(LT\right)}=n_{SO_2}=0,6\left(mol\right)\\ n_{S\left(TT\right)}=0,6:75\%=0,8\left(mol\right)\\ m_{S\left(TT\right)}=0,8.32=25,6\left(g\right)\)

S+O2-to>SO2

0,6-----------0,6 mol

n SO2=13,44\22,4=0,6 mol

=>H=75%

=>m S=0,6.32.100\75=25,6g

\(n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\\ n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\\ 4P+5O_2\underrightarrow{^{to}}2P_2O_5\\ Vì:\dfrac{0,6}{5}>\dfrac{0,2}{1}\\ \Rightarrow O_2dư\\ n_{P_2O_5\left(LT\right)}=\dfrac{2}{4}.0,2=0,1\left(mol\right)\\ n_{P_2O_5\left(TT\right)}=0,1.75\%=0,075\left(mol\right)\\ m_{P_2O_5\left(TT\right)}=142.0,075=10,65\left(g\right)\)

\(n_P=\dfrac{6,2}{31}=0,2\left(mol\right);n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

PTHH: 4P + 5O₂ ---t^o→ 2P₂O₅

Mol:     0,2                        0,1

Ta có:\(\dfrac{0,2}{4}< \dfrac{0,6}{5}\) ⇒ P hết, O₂ dư

\(m_{P_2O_5\left(lt\right)}=0,1.142=14,2\left(g\right)\Rightarrow m_{P_2O_5\left(tt\right)}=\dfrac{14,2}{75}.100=18,94\left(g\right)\)

\(Đặt:n_{KMnO_4\left(LT\right)}=a\left(mol\right)\\ 2KMnO_4\underrightarrow{^{to}}K_2MnO_4+MnO_2+O_2\\ Vì:m_{rắn}=29,04\\ \Leftrightarrow\left(31,6-158a\right)+197.0,5a+87.0,5a=29,04\\ \Leftrightarrow a=0,16\\ \Rightarrow H=\dfrac{0,16.158}{31,6}.100=80\%\)

\(Đặt:n_{KMnO_4\left(LT\right)}=a\left(mol\right)\\ 2KMnO_4\underrightarrow{^{to}}K_2MnO_4+MnO_2+O_2\\ n_{KMnO_4\left(bđ\right)}=\dfrac{31,6}{158}=0,2\left(mol\right)\\ n_{KMnO_4\left(LT\right)}=0,2-a\left(mol\right)\\ n_{K_2MnO_4}=n_{MnO_2}=0,5a\left(mol\right)\\ m_{rắn}=29,04\\ \Leftrightarrow m_{KMnO_4\left(LT\right)}+m_{K_2MnO_4}+m_{MnO_2}=29,04\\ \Leftrightarrow\left(31,6-158a\right)+197a.0,5+87a.0,5=29,04\\ \Leftrightarrow a=0,16\)

\(\Rightarrow H=\dfrac{0,16}{0,2}.100=80\%\)

a. \(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)

\(n_{KMnO_4}=0,6mol\)

\(\rightarrow n_{O_2}=\frac{1}{2}n_{KMnO_4}=0,3mol\)

\(\rightarrow V_{O_2}=6,72l\)

\(V_{O_2\text{thực}}=\frac{6,72.75}{100}=5,04l\)

b. \(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)

\(n_{O_2}=1,5mol\)

\(\rightarrow n_{KMnO_4}=2n_{O_2}=3mol\)

\(\rightarrow m_{KMnO_4\text{cần}}=\frac{474.100}{80}=592,5g\)

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

S+O2-to>SO2

0,2----------0,2 mol

n S=6,4\32=0,2 mol

H=80%

VSO2=0,2.22,4.80\100=3.584l