`PTHH:2KClO_3` $\xrightarrow[]{t^o}$ `2KCl+3O_2`

Theo bài ra: `n_{KCl}=(2,086)/(74,5)=0,028(mol)`

Theo PTHH: `n_{KClO_3(t\t)}=n_{KCl}=0,028(mol)`

`=>m_{KClO_3(t\t)}=0,028.122,5=3,43(g)`

`a,H=(3,43)/(4,9).100\%=70\%`

`b,` Theo PTHH: `n_{O_2}=1,5n_{KCl}=1,5.0,028=0,042(mol)`

`V_{O_2(đkc)}=0,042.24,79=1,04118(l)`

`a)`

`n_{KClO_3}={4,9}/{122,5}=0,04\ (mol)`

`n_{KCl\ t t}={2,086}/{74,5}=0,028\ (mol)`

`2KClO_3->_{MnO_2}^{t^o}2KCl+3O_2`

Theo PT: `n_{KCl\ l t}=n_{KClO_3}=0,04\ (mol)`

`->H={0,028}/{0,04}.100%=70%`

`b)`

Theo PT: `n_{O_2}={3}/{2}n_{KCl}=0,042\ (mol)`

`->V_{O_2}=0,042.24,79=1,04118\ (L)`