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

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

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

đây nè bạn!

\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\\ Đặt:n_{CuO\left(p.ứ\right)}=a\left(mol\right)\\ \Rightarrow n_{Cu}=a\left(mol\right);m_{CuO\left(dư\right)}=24-80a\left(g\right)\\ \Rightarrow m_{rắn}=m_{CuO\left(dư\right)}+m_{Cu}=\left(24-80a\right)+64a=21,6\\ \Leftrightarrow-16a=-2,4\\ \Leftrightarrow a=0,15\\ Vậy:H=\dfrac{0,15.80}{24}.100\%=50\%\\ b,n_{H_2}=n_{Cu}=a=0,15\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\)

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

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%

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

CuO+H2--->Cu+H2O

Chất rắn k tan là Cu

n CuO=12/80=0,15(mol)

n Cu=6,6/64=0,103125(mol)

-->Cuo dư

n CuO=n Cu=0,103125(mol)

H=0,103125/0,15.100%=68,75%

Vì Cu không tác dụng với HCl nên ta có chất rắn không tan là Cu

Theo−đề−bài−ta−có:Theo−đề−bài−ta−có: nCuO = 1280=0,15(mol)1280=0,15(mol) ; nCu= 6,664≈0,103(mol)6,664≈0,103(mol)

Ta có PTHH :

CuO + H2−t0→Cu+H2O−t0→Cu+H2O

Theo PTHH ta có :

nCuO=0,151mol>nCu=0,1031molnCuO=0,151mol>nCu=0,1031mol nên => nCuO dư ; nCu hết

Ta có :

H=(số−mol−chất−thiếu)(số−mol−chất−dư).100%=0,1030,15.100%≈68,67%H=(số−mol−chất−thiếu)(số−mol−chất−dư).100%=0,1030,15.100%≈68,67%

Vậy ...............

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