Bài 3:
a, Ta có: \(n_{KClO_3}=\dfrac{4,9}{122,5}=0,04\left(mol\right)\)
PT: \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
_____0,04___________0,06 (mol)
\(\Rightarrow V_{O_2}=0,06.22,4=1,344\left(l\right)\)
b, Ta có: \(n_{H_2O}=\dfrac{3,6.1000}{18}=200\left(mol\right)\)
PT: \(2H_2O\underrightarrow{đp}2H_2+O_2\)
____200_________100 (mol)
\(\Rightarrow V_{O_2}=100.22,4=2240\left(l\right)\)
Bạn tham khảo nhé!
2KClO3-to>2KCl+3O2
0,04------------------0,06 mol
n KClO3=\(\dfrac{4,9}{122,5}\)=0,04 mol
=>VO2=0,06.22,4=1,344 l
b) 2H2O-đp->2H2+O2
0,2-----------------------0,1 kmol
n H2O=\(\dfrac{3,6}{18}\)=0,2 kmol
=>VO2=100.22,4=2240l
Bài 3:
\(a,PTHH:2KClO_3\underrightarrow{t^o}2KCl+3O_2\\ n_{KClO_3}=\dfrac{m}{M}=\dfrac{4,9}{122,5}=0,04\left(mol\right)\\ Theo.PTHH:n_{O_2}=\dfrac{3}{2}.n_{KClO_3}=\dfrac{3}{2}.0,04=0,06\left(mol\right)\\ V_{O_2}=n.22,4=0,06.22,4=1,344\left(l\right)\)
\(b,Đổi:3,6Kg=3600g\\ n_{H_2O}=\dfrac{m}{M}=\dfrac{3600}{18}=200\left(mol\right)\\ PTHH:2H_2O\underrightarrow{điện.phân}2H_2+O_2\\ Theo.PTHH:n_{O_2}=\dfrac{1}{2.}n_{H_2O}=\dfrac{1}{2}.200=100\left(mol\right)\\ V_{O_2\left(đktc\right)}=n.22,4=100.22,4=2240\left(l\right)\)
