Question #fca37

1 Answer
May 20, 2017

Answer:

#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#