a) \(2KClO_3\underrightarrow{t^0}2KCl+3O_2\)
b) \(n_{KClO_3}=\frac{142}{122,5}=1,15\left(mol\right)\)
Theo PTHH: \(n_{KClO_3}:n_{O_2}=2:3\)
\(\Rightarrow n_{O_2}=n_{KClO_3}.\frac{3}{2}=1,15.\frac{3}{2}=1,725\left(mol\right)\)
\(\Rightarrow V_{O_2\left(đktc\right)}=1,725.22,4=38,64\left(l\right)\)
c) Theo PTHH: \(n_{KClO_3}:n_{KCl}=2:2=1\)
\(\Rightarrow n_{KCl}=n_{KClO_3}=1,15\left(mol\right)\)
\(\Rightarrow m_{KCl}=1,15.74,5=85,675\left(g\right)\)
a)nKClO3=142/122.5=1.2(mol) (mik để xấp xỉ nha)
PTHH: 2KClO3 --to-->2KCl+3O2
theo pt: 2 2 3 (mol)
theo db: 1.2 1.2 1.8 (mol) (mik làm tắt)
b)VO2 =1.8*22.4=40.32(l)
c)mKCl=1.2*74.5=89.4(g)