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%

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

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

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

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

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

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

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%

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 ---t^o→ K₂MnO₄ + MnO₂ + O₂

Mol:       1,2                                                    0,6

\(m_{KMnO_4\left(lt\right)}=1,2.158=189,6\left(g\right)\Rightarrow m_{KMnO_4\left(tt\right)}=\dfrac{189,6}{75}.100=252,8\left(g\right)\)

a) \(n_{H_2O}=\dfrac{2,4.10^{23}}{6.10^{23}}=0,4\left(mol\right)\)

=> \(n_{H_2\left(pư\right)}=0,4\left(mol\right)\)

Theo ĐLBTKL

=> \(m=28,4+0,4.18-0,4.2=34,8\left(g\right)\)

b) \(n_{Fe\left(X\right)}=\dfrac{28,4.59,155\%}{56}=0,3\left(mol\right)\)

n_O = n_H2O = 0,4 (mol)

=> n_Fe : n_O = 3:4

=> CTHH: Fe₃O₄

c) \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)

 \(n_{Fe_3O_4}=\dfrac{34,8}{232}=0,15\left(mol\right)\)

\(n_{H_2\left(bd\right)}=\dfrac{17,92}{22,4}=0,8\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,15}{1}< \dfrac{0,8}{4}\) => Hiệu suất tính theo Fe₃O₄

n_Fe(X) = 0,3 (mol)

=> n_{Fe3O4 (bị khử)} = 0,1 (mol)

=> \(\dfrac{0,1}{0,15}.100\%=66,67\%\)

\(n_{H_2}=\dfrac{17,92}{22,4}=0,8\left(mol\right)\\ n_{H_2O}=\dfrac{2,4.10^{23}}{6.10^{23}}=0,4\left(mol\right)\\ Đặt.oxit.sắt:Fe_xO_y\left(x,y:nguyên,dương\right)\\ Fe_xO_y+yH_2\rightarrow\left(t^o\right)xFe+yH_2O\\ Vì:\dfrac{0,4}{y}< \dfrac{0,8}{y}\\ \Rightarrow H_2dư\\ \Rightarrow n_{H_2\left(p.ứ\right)}=n_{H_2O}=n_{O\left(mất\right)}=0,4\left(mol\right)\\ a,m=m_{oxit}=m_{rắn}+m_O=28,4+0,4.16=34,8\left(g\right)\\b,m_{Fe}=28,4.59,155\%=16,8\left(g\right)\\ \Rightarrow n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\\ \Rightarrow x:y=0,3:0,4=3:4\\ \Rightarrow CTHH:Fe_3O_4\\ c,n_{Fe_3O_4\left(bđ\right)}=\dfrac{34,8}{232}=0,15\left(mol\right)\\ \Rightarrow n_{Fe\left(LT\right)}=3.0,15=0,45\left(mol\right)\\ n_{Fe\left(TT\right)}=0,3\left(mol\right)\)

\(\Rightarrow H=\dfrac{0,3}{0,45}.100=66,667\%\)

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

\(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\\ n_{Cu\left(TT\right)}=\dfrac{10,24}{64}=0,16\left(mol\right)\\ CuO+H_2\underrightarrow{^{to}}Cu+H_2O\\ n_{Cu\left(LT\right)}=n_{CuO}=0,2\left(mol\right)\\ H=\dfrac{0,16}{0,2}.100=80\%\)