Answer
a) $n(O_{2})=0.294mol$
b) $n(O_{2})=0.353mol$
c) $n(O_{2})=0.055mol$
Work Step by Step
a) $n(KClO_{3}):n(O_{2})=2:3 \implies n(O_{2})=\frac{3}{2}n(KClO_{3})=1.5\times \frac{m(KClO_{3})}{M(KClO_{3})}=1.5 \times \frac{24g}{122.5\frac{g}{mol}}=0.294mol$
b) $n(H_{2}O_{2}):n(O_{2})=2:1 \implies n(O_{2})=\frac{1}{2}n(H_{2}O_{2})=0.5\times \frac{m(H_{2}O_{2})}{M(H_{2}O_{2})}=0.5 \times \frac{24g}{34\frac{g}{mol}}=0.353mol$
c) $n(HgO):n(O_{2})=2:1 \implies n(O_{2})=\frac{1}{2}n(HgO)=0.5\times \frac{m(HgO)}{M(HgO)}=0.5 \times \frac{24g}{217\frac{g}{mol}}=0.055mol$
