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\(n_{KClO_3}=\dfrac{m_{KClO_3}}{M_{KClO_3}}=\dfrac{15}{122,5}=\dfrac{6}{49}mol\)
\(n_{KClO_3}=\dfrac{6}{49}:90\%=\dfrac{20}{147}mol\)
\(2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\)
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 --to--> 2KCl + 3O2
\(\dfrac{27}{245}\)----------------->\(\dfrac{81}{490}\)
=> \(V_{O_2}=\dfrac{81}{490}.22,4=\dfrac{648}{175}\left(l\right)\)
Ta có: \(n_{O_2}=\dfrac{8,4}{22,4}=0,375\left(mol\right)\)
PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
Theo PT: \(n_{KMnO_4\left(LT\right)}=2n_{O_2}=0,75\left(mol\right)\)
Mà: H = 80% \(\Rightarrow n_{KMnO_4\left(TT\right)}=\dfrac{0,75}{80\%}=0,9375\left(mol\right)\)
\(\Rightarrow m_{KMnO_4\left(TT\right)}=0,9375.158=148,125\left(g\right)\)
Mà: Thuốc tím chứa 20% tạp chất trơ.
\(\Rightarrow a=\dfrac{148,125}{80\%}=185,15625\left(g\right)\)
\(a,2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2\\ n_{KMnO_4}=\dfrac{395}{158}=2,5(mol)\\ \Rightarrow n_{O_2}=1,25(mol)\\ \Rightarrow V_{O_2}=1,25.22,4=28(l)\\ \Rightarrow V_{O_2(tt)}=28.85\%=23,8(l)\)
\(b,n_{O_2}=\dfrac{67,2}{22,4}=3(mol)\\ 2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2\\ \Rightarrow n_{KMnO_4}=6(mol)\\ \Rightarrow m_{KMnO_4}=6.158=948(g)\\ \Rightarrow m_{KMnO_4(tt)}=\dfrac{948}{80\%}=1185(g)\)
\(n_{O_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
PTHH: 2KMnO4 --to--> K2MnO4 + MnO2 + O2
1<------------------------------------0,5
=> \(m_{KMnO_4\left(pthh\right)}=1.158=158\left(g\right)\)
=> \(m_{KMnO_4\left(tt\right)}=\dfrac{158.100}{80}=197,5\left(g\right)\)
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%
\(\rightarrow V_{O_2}=0,0115.22,4.80\%=0,20608l\)
\(n_{KMnO_4}=\dfrac{63,2}{158}=0,4\left(mol\right)\\
pthh:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
0,4 0,2
=> \(V_{O_2\left(lt\right)}=0,2.22,4=4,48\left(l\right)\\
V_{O_2\left(tt\right)}=\dfrac{90.4,48}{100}=4,032\left(l\right)\)
\(n_{KClO_3\left(bd\right)}=\dfrac{55,125}{122,5}=0,45\left(mol\right)\)
=> \(n_{KClO_3\left(pư\right)}=\dfrac{0,45.85}{100}=0,3825\left(mol\right)\)
PTHH: 2KClO3 --to,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%
=>VO2=0,675.22,4.\(\dfrac{85}{100}\)=12,852l
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. \(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)
\(n_{O_2}=1,5mol\)
\(\rightarrow n_{KMnO_4}=2n_{O_2}=3mol\)
\(\rightarrow m_{KMnO_4\text{cần}}=\frac{474.100}{80}=592,5g\)