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

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

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

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

\(n_{KClO_3\left(bd\right)}=\dfrac{55,125}{122,5}=0,45\left(mol\right)\)

=> \(n_{KClO_3\left(pư\right)}=\dfrac{0,45.85}{100}=0,3825\left(mol\right)\)

PTHH: 2KClO₃ --t^o,MnO₂--> 2KCl + 3O₂

             0,3825------------------->0,57375

=> \(V_{O_2}=0,57375.22,4=12,852\left(l\right)\)

2KClO₃-to>2KCl+3O₂

0,45---------------------0,675 mol

n _KClO3=\(\dfrac{55,125}{122,5}\)=0,45 mol

=>H=85%

=>V_O2=0,675.22,4.\(\dfrac{85}{100}\)=12,852l

\(a,2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2\\ n_{KMnO_4}=\dfrac{395}{158}=2,5(mol)\\ \Rightarrow n_{O_2}=1,25(mol)\\ \Rightarrow V_{O_2}=1,25.22,4=28(l)\\ \Rightarrow V_{O_2(tt)}=28.85\%=23,8(l)\)

\(b,n_{O_2}=\dfrac{67,2}{22,4}=3(mol)\\ 2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2\\ \Rightarrow n_{KMnO_4}=6(mol)\\ \Rightarrow m_{KMnO_4}=6.158=948(g)\\ \Rightarrow m_{KMnO_4(tt)}=\dfrac{948}{80\%}=1185(g)\)

\(n_{CaCO_3}=\dfrac{60}{100}=0,6mol\)

\(n_{CaCO_3}=0,6.80\%=0,48mol\)

\(CaCO_3\rightarrow\left(t^o\right)CaO+CO_2\)

  0,48                             0,48 ( mol )

\(V_{CO_2}=0,48.22,4=10,752l\)

=>Chọn C

\(n_{KClO_3}=\dfrac{m_{KClO_3}}{M_{KClO_3}}=\dfrac{15}{122,5}=\dfrac{6}{49}mol\)

\(n_{KClO_3}=\dfrac{6}{49}:90\%=\dfrac{20}{147}mol\)

\(2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\)

 20/147                          10/49   ( mol )

\(V_{O_2}=n_{O_2}.22,4=\dfrac{10}{49}.22,4=4,5714l\)

\(m_{KClO_3\left(pư\right)}=\dfrac{15.90}{100}=13,5\left(g\right)\)

=> \(n_{KClO_3\left(pư\right)}=\dfrac{13,5}{122,5}=\dfrac{27}{245}\left(mol\right)\)

PTHH: 2KClO₃ --t^o--> 2KCl + 3O₂

            \(\dfrac{27}{245}\)----------------->\(\dfrac{81}{490}\)

=> \(V_{O_2}=\dfrac{81}{490}.22,4=\dfrac{648}{175}\left(l\right)\)

a. PTHH: \(KMnO_4\rightarrow^{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)

b. \(H=100\%\)

\(n_{KMnO_4}=\frac{3,6}{158}=0,023mol\)

Theo phương trình \(n_{O_2}=0,5n_{KMnO_4}=0,046mol\)

\(\rightarrow V_{O_2}=0,0115.22,4.100\%=0,2576l\)

c. H = 80%

\(\rightarrow V_{O_2}=0,0115.22,4.80\%=0,20608l\)

\(n_{KMnO_4}=\dfrac{79}{158}=0,5mol\)

\(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

0,5                                                 0,25

\(H=80\%\Rightarrow n_{O_2}=0,25\cdot80\%=0,2mol\)

\(\Rightarrow V=0,2\cdot22,4=4,48l\)

PT: \(2KMnO_4+16HCl\rightarrow2MnCl_2+2KCl+5Cl_2+8H_2O\)

Ta có: \(n_{KMnO_4}=\dfrac{31,6}{158}=0,2\left(mol\right)\)

\(m_{HCl}=\dfrac{60.36,5}{100}=21,9\left(g\right)\Rightarrow n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,2}{2}>\dfrac{0,6}{16}\), ta được KMnO₄ dư.

Theo PT: \(n_{Cl_2\left(LT\right)}=\dfrac{5}{16}n_{HCl}=0,1875\left(mol\right)\)

Mà: H% = 80%

\(\Rightarrow n_{Cl_2\left(TT\right)}=0,1875.80\%=0,15\left(mol\right)\)

\(\Rightarrow V_{Cl_2}=0,15.22,4=3,36\left(l\right)\)

Bạn tham khảo nhé!