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\(Đặt:n_{KMnO_4\left(LT\right)}=a\left(mol\right)\\ 2KMnO_4\underrightarrow{^{to}}K_2MnO_4+MnO_2+O_2\\ n_{KMnO_4\left(bđ\right)}=\dfrac{31,6}{158}=0,2\left(mol\right)\\ n_{KMnO_4\left(LT\right)}=0,2-a\left(mol\right)\\ n_{K_2MnO_4}=n_{MnO_2}=0,5a\left(mol\right)\\ m_{rắn}=29,04\\ \Leftrightarrow m_{KMnO_4\left(LT\right)}+m_{K_2MnO_4}+m_{MnO_2}=29,04\\ \Leftrightarrow\left(31,6-158a\right)+197a.0,5+87a.0,5=29,04\\ \Leftrightarrow a=0,16\)
\(\Rightarrow H=\dfrac{0,16}{0,2}.100=80\%\)
\(n_{O_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\\ n_{KMnO_4}=\dfrac{15,8}{158}=0,1\left(mol\right)\\ 2KMnO_4\underrightarrow{^{to}}K_2MnO_4+MnO_2+O_2\\ n_{O_2\left(LT\right)}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ H=\dfrac{0,03}{0,05}.100=60\%\)
\(Đặt:n_{KClO_3\left(LT\right)}=a\left(mol\right)\\ 2KClO_3\underrightarrow{^{to}}2KCl+3O_2\\ n_{KCl}=a\left(mol\right)\\ m_{rắn}=30,99\\ \Leftrightarrow\left(36,75-122,5a\right)+74,5a=30,99\\ \Leftrightarrow a=0,12\\ m_{KClO_3\left(LT\right)}=0,12.122,5=14,7\left(g\right)\\ H=\dfrac{14,7}{36,75}.100=40\%\)
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 ---to→ K2MnO4 + MnO2 + O2
Mol: 1,2 0,6
\(m_{KMnO_4\left(lt\right)}=1,2.158=189,6\left(g\right)\Rightarrow m_{KMnO_4\left(tt\right)}=\dfrac{189,6}{75}.100=252,8\left(g\right)\)
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%
\(n_{O_2\left(TT\right)}=\dfrac{2,016}{22,4}=0,09\left(mol\right)\\ n_{KClO_3}=\dfrac{12,25}{122,5}=0,1\left(mol\right)\\ 2KClO_3\underrightarrow{^{to}}2KCl+3O_2\\ n_{O_2\left(LT\right)}=\dfrac{3}{2}.0,1=0,15\left(mol\right)\\ \Rightarrow H=\dfrac{0,09}{0,15}.100=60\%\)
\(n_{KClO_3}=\dfrac{24,5}{122,5}=0,2\left(mol\right)\\ 2KClO_3\underrightarrow{^{to}}2KCl+3O_2\\ n_{O_2\left(LT\right)}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\\ Vì:H=90\%\Rightarrow n_{O_2\left(TT\right)}=90\%.0,3=0,27\left(mol\right)\\ V_{O_2\left(đktc,thực.tế\right)}=0,27.22,4=6,048\left(l\right)\)
\(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\\ n_{Cu\left(TT\right)}=\dfrac{10,24}{64}=0,16\left(mol\right)\\ CuO+H_2\underrightarrow{^{to}}Cu+H_2O\\ n_{Cu\left(LT\right)}=n_{CuO}=0,2\left(mol\right)\\ H=\dfrac{0,16}{0,2}.100=80\%\)
a)
$2Cu(NO_3)_2 \xrightarrow{t^o} 2CuO + 4NO_2 + O_2$
Theo PTHH :
Gọi $n_{CuO} = n_{Cu(NO_3)_2\ pư} = a(mol)$
Ta có :
$m_{Chất\ rắn} = 80a + 20 - 188a = 9,2 \Rightarrow a = 0,1$
$H = \dfrac{0,1.188}{20}.100\% = 94\%$
b)
Theo PTHH :
$n_{NO_2} = 2n_{CuO} = 0,2(mol)$
$n_{O_2} = \dfrac{1}{2}n_{CuO} = 0,05(mol)$
$V = (0,2 + 0,05).22,4 = 5,6(lít)$
\(n_{O_2}=a\left(mol\right)\)
\(2Cu\left(NO_3\right)_2\underrightarrow{^{^{t^0}}}2CuO+4NO_2+O_2\)
\(2a..............2a.........4a...a\)
\(BTKL:\)
\(m_{khí}=20-9.2=10.8\left(g\right)\)
\(\Leftrightarrow4a\cdot46+32a=10.8\)
\(\Leftrightarrow a=0.05\)
\(H\%=\dfrac{0.05\cdot2\cdot188}{20}\cdot100\%=94\%\)
\(V=\left(0.05+0.05\cdot4\right)\cdot22.4=5.6\left(l\right)\)
\(Đặt:n_{KMnO_4\left(LT\right)}=a\left(mol\right)\\ 2KMnO_4\underrightarrow{^{to}}K_2MnO_4+MnO_2+O_2\\ Vì:m_{rắn}=29,04\\ \Leftrightarrow\left(31,6-158a\right)+197.0,5a+87.0,5a=29,04\\ \Leftrightarrow a=0,16\\ \Rightarrow H=\dfrac{0,16.158}{31,6}.100=80\%\)