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a, PT: \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
Ta có: \(n_{KCl}=\dfrac{1,49}{74,5}=0,02\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{3}{2}n_{KCl}=0,03\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,03.24,79=0,7437\left(l\right)\)
b, Theo PT: \(n_{KClO_3\left(TT\right)}=n_{KCl}=0,02\left(mol\right)\)
\(\Rightarrow m_{KClO_3\left(TT\right)}=0,02.122,5=2,45\left(g\right)\)
\(\Rightarrow H=\dfrac{2,45}{3,5}.100\%=70\%\)
2KClO3 \(\underrightarrow{to}\) 2KCl + 3O2
\(n_{KCl}=\frac{2,5}{74,5}=\frac{5}{149}\left(mol\right)\)
Theo pT: \(n_{KClO_3}pư=n_{KCl}=\frac{5}{149}\left(mol\right)\)
\(\Rightarrow m_{KClO_3}pư=\frac{5}{149}\times122,5=4,11\left(g\right)\)
\(H\%=\frac{4,11}{4,9}\times100\%=83,88\%\)
Có :\(n_{KClO_3}=\frac{4,9}{122,5}=0,04mol\)
PTHH : \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\left(1\right)\)
(mol) 0,04 \(\rightarrow\) 0,04
Từ (1) => \(n_{KCl}=0,04mol\Rightarrow m_{KCl}=0,04.74,5=2,98g\)
H = \(\frac{2,5}{2,98}.100\%\approx84\%\)
\(a.2KClO_3\xrightarrow[t^0]{xt}2KCl+3O_2\\ b.n_{KClO_3}=\dfrac{m_{KClO_3}}{M_{KClO_3}}=\dfrac{22,05}{122,5}=0,18mol\\ n_{O_2\left(lt\right)}=\dfrac{0,18.3}{2}=0,27mol\\ V_{O_2\left(lt\right)}=n_{O_2\left(lt\right)}.22,4=0,27.22,4=6,048l\\ H=\dfrac{V_{O_2\left(tt\right)}}{V_{O_2\left(lt\right)}}\cdot100\%=\dfrac{3,36}{6,048}\cdot100\%\approx55,56\%\\ c.n_{O_2}=\dfrac{V_{O_2}}{22,4}=\dfrac{3,36}{22,4}=0,15mol\\ n_{KCl}=\dfrac{0,15.2}{3}=0,1mol\\ m_{KCl}=n_{KCl}.M_{KCl}=0,1.74,5=7,45g\)
\(n_{KClO_3}=\dfrac{7}{122,5}=\dfrac{2}{35}\left(mol\right)\)
PTHH :
\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
2/35 2/35 3/35
\(V_{O_2}=\dfrac{3}{35}.24,79\approx2,1249\left(l\right)\)
\(m_{KCl}=\dfrac{2}{35}.74,5\approx4,257\left(g\right)\)
\(H=\dfrac{2,98}{4,257}.100\%=70\%\)
Gọi hiệu suất phản ứng là a
$2SO_2 + O_2 \xrightarrow{t^o,V_2O_5} 2SO_3$
Ta thấy :
V SO2 / 2 = V O2 nên hiệu suất tính theo số mol của SO2 hoặc O2
V SO2 phản ứng = 4,48a(lít)
V O2 phản ứng = 2,24a(lít)
V SO3 = V SO2 pư = 4,48a(lít)
Sau phản ứng, khí gồm :
SO2 : 4,48 - 4,48a(lít)
O2 : 2,24 - 2,24a(lít)
SO3 : 4,48a(lít)
Suy ra :
4,48 - 4,48a + 2,24 - 2,24a + 4,48a = 5,6
=> a = 0,5 = 50%
Hỗn hợp khí gồm :
SO2 : 4,48 -4,48.0,5 = 2,24 lít
O2 : 2,24 - 2,24.0,5 = 1,12 lít
SO3 : 4,48.0,5 = 2,24 lít
\(a,PTHH:2KClO_3\xrightarrow{t^o}2KCl+3O_2\\ b,n_{KClO_3(thực tế)}=\dfrac{36,75}{122,5}=0,3(mol)\\ n_{O_2(phản ứng)}=\dfrac{6,72}{22,4}=0,3(mol)\\ \Rightarrow n_{KClO_3(phản ứng)}=\dfrac{2}{3}n_{O_2}=0,2(mol)\\ \Rightarrow H\%=\dfrac{0,2}{0,3}.100\%=66,67\%\\ c,n_{KCl}=n_{KClO_3(phản ứng)}=0,2(mol)\\ \Rightarrow m_{KCl}=0,2.74,5=14,9(g)\)
\(n_{KClO_3\left(bđ\right)}=\dfrac{24,5}{122,5}=0,2\left(mol\right)\)
=> \(n_{KClO_3\left(pư\right)}=\dfrac{0,2.50}{100}=0,1\left(mol\right)\)
PTHH: 2KClO3 --to--> 2KCl + 3O2
0,1------------------>0,15
=> V = 0,15.22,4 = 3,36 (l)
nKClO3 = 24,5/122,5 = 0,2 (mol)
PTHH: 2KClO3 -> (t°, MnO2) 2KCl + 3O2
Mol: 0,2 ---> 0,2 ---> 0,3
nO2 (TT) = 0,3 . 50% = 0,15 (mol)
VO2 (TT) = 0,15 . 22,4 = 3,36 (l)
\(n_{KClO3}=\dfrac{5,25}{122,5}=\dfrac{3}{70}\left(mol\right)\)
a) PTHH : \(2KClO_3\xrightarrow[]{t^o}2KCl+3O_2\)
\(\dfrac{3}{70}\) \(\dfrac{3}{70}\) \(\dfrac{9}{140}\)
b) \(V_{O2\left(dktc\right)}=\dfrac{9}{140}.22,4=1,44\left(l\right)\)
c) \(m_{KCl\left(lt\right)}=\dfrac{3}{70}.74,5=\dfrac{447}{140}\left(mol\right)\)
\(\Rightarrow H\%=\dfrac{m_{tt}}{m_{lt}}.100\%=\dfrac{2,235}{\dfrac{447}{140}}.100\%=70\%\)
nKClO3=0,04 mol
nKCl=0,034 mol
2KClO3. =>2KCl. +3O2
0,034 mol<=0,034 mol=>0,051 mol
H%=0,034/0,04.100%=83,89%
VO2=0,051.22,4=1,1424 lit
\(n_{KClO_3}=\frac{4,9}{122,5}=0,04\left(mol\right)\)
\(n_{KCl}=\frac{2,5}{74,5}=0,034\left(mol\right)\)
\(2KClO_3->2KCl+3O_2\left(1\right)\)
theo (1) \(n_{KClO_3\left(pư\right)}=n_{KCl}=0,034\left(mol\right)\)
hiệu suất phản ứng là
\(\frac{0,034}{0,04}.100\%=83,89\%\)
theo (1) \(n_{O_2}=\frac{3}{2}n_{KCl}=0,051\left(mol\right)\)
=> \(V_{O_2}=0,051.22,4=1,1424\left(l\right)\)