$2\text{KC}l{\text{O}}_3\left(\text{s}\right)\to 2\text{KC}l\left(\text{s}\right)+3{\text{O}}_{\text{2}}\left(\text{g}\right)$ from the above the reaction match the following items

A) 4 moles of gives	1) 2 KCl
B) 2 moles of gives	2) $3{\text{\hspace{0.17em}O}}_2$
C) 2 moles of gives	3) 4 KCl

From the stoichiometric equation, the amount of products obtained were given below: 

   $2\text{KC}l{\text{O}}_3\left(\text{s}\right)\to 2\text{KC}l\left(\text{s}\right)+3{\text{O}}_{\text{2}}\left(\text{g}\right)$

$\text{4\hspace{0.17em}KC}l{\text{O}}_3\left(\text{s}\right)\to 4\text{\hspace{0.17em}KC}l\left(\text{s}\right)+6{\text{\hspace{0.17em}O}}_{\text{2}}\left(\text{g}\right)$

$4{\text{ mole KClO}}_3=4\text{ mole KCl}$

${\text{2 mole KClO}}_3=\text{2 mole KCl}$

${\text{2 mole KClO}}_3=3{\text{\hspace{0.17em}mole O}}_{\text{2}}$