Answer
a. 18.0 g of $O_2$
b. 67.2 %
Work Step by Step
a.
$ KClO_3 $ : ( 35.45 $\times$ 1 )+ ( 16.00 $\times$ 3 )+ ( 39.10 $\times$ 1 )= 122.55 g/mol
$$ \frac{1 \space mole \space KClO_3 }{ 122.55 \space g \space KClO_3 } \space and \space \frac{ 122.55 \space g \space KClO_3 }{1 \space mole \space KClO_3 }$$
$ O_2 $ : ( 16.00 $\times$ 2 )= 32.00 g/mol
$$ \frac{1 \space mole \space O_2 }{ 32.00 \space g \space O_2 } \space and \space \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 }$$
$$ 46.0 \space g \space KClO_3 \times \frac{1 \space mole \space KClO_3 }{ 122.55 \space g \space KClO_3 } \times \frac{ 3 \space moles \space O_2 }{ 2 \space moles \space KClO_3 } \times \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 } = 18.0 \space g \space O_2 $$
b.
Calculation of percent yield: $$\frac{ 12.1 \space g \space O_2 }{ 18.0 \space g \space O_2 } \times 100\% = 67.2 \%$$
