Question

Question #fca37

Answer 1

`4.1 gram`.

Explanation:

Decomposition of `KClO_3` : `2 KClO_3(s) → 3 O_2(g) + 2 KCl(s)`
Decomposition of `HgO`: `2HgO(s)→2Hg(l)+O_2(g)`

Moles of oxygen produced from the decomposition of mercury II oxide:
`(21.6 *g HgO)/(216 g*mol^-1 HgO)* (1 mol O_2)/(2 mol HgO)=0.05 mol O_2`
Moles of potassium chlorite required:
`(0.05* mol O_2)*(2 mol KClO_3)/(3 *mol O_2)=0.033mol KClO_3`
Mass of potassium chlorite required:
`0.033* mol KClO_3 * 123* g*mol^-1 KClO_3 = 4.1 g`