\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\\ CuO+H_2\underrightarrow{^{to}}Cu+H_2O\\ Vì:\dfrac{0,3}{1}>\dfrac{0,2}{1}\\ \Rightarrow H_2dư\\ n_{Cu\left(LT\right)}=n_{CuO}=0,2\left(mol\right)\\ n_{Cu\left(TT\right)}=\dfrac{8,96}{64}=0,14\left(mol\right)\\ H=\dfrac{0,14}{0,2}.100=70\%\)

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

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

\(Đặt:n_{KClO_3\left(LT\right)}=a\left(mol\right)\\ 2KClO_3\underrightarrow{^{to}}2KCl+3O_2\\ n_{KCl}=a\left(mol\right)\\ m_{rắn}=30,99\\ \Leftrightarrow\left(36,75-122,5a\right)+74,5a=30,99\\ \Leftrightarrow a=0,12\\ m_{KClO_3\left(LT\right)}=0,12.122,5=14,7\left(g\right)\\ H=\dfrac{14,7}{36,75}.100=40\%\)

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%

\(n_{O_2\left(TT\right)}=\dfrac{2,016}{22,4}=0,09\left(mol\right)\\ n_{KClO_3}=\dfrac{12,25}{122,5}=0,1\left(mol\right)\\ 2KClO_3\underrightarrow{^{to}}2KCl+3O_2\\ n_{O_2\left(LT\right)}=\dfrac{3}{2}.0,1=0,15\left(mol\right)\\ \Rightarrow H=\dfrac{0,09}{0,15}.100=60\%\)

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

n HCl  = 0,4 mol

2Al + 6HCl → 2Al Cl 3  + 3 H 2 (1)

H 2  + CuO  → t ° Cu +  H 2 O(2)

Theo (1) và (2) ta có sơ đồ chuyển hóa:

6 mol HCl → 3 mol  H 2  → 3 mol Cu

0,4 mol HCl       x mol Cu

x = 0,4 x 3 /6 = 0,2 mol → m Cu  = 0,2 x 64 = 12,8g

H% = 11,52/12,8 . 100% = 90%

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

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%

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