Answer
(a) Theoretical Yield: 7.05 g of $O_2$
(b) Percent Yield: 92.9%
Work Step by Step
1. Find the molar masses:
$ C_3H_5N_3O_9 $ : ( 1.008 $\times$ 5 )+ ( 12.01 $\times$ 3 )+ ( 14.01 $\times$ 3 )+ ( 16.00 $\times$ 9 )= 227.10 g/mol
$$ \frac{1 \space mole \space C_3H_5N_3O_9 }{ 227.10 \space g \space C_3H_5N_3O_9 } \space and \space \frac{ 227.10 \space g \space C_3H_5N_3O_9 }{1 \space mole \space C_3H_5N_3O_9 }$$
$ O_2 $ : ( 16.00 $\times$ 2 )= 32.00 g/mol
$$ \frac{1 \space mole \space O_2 }{ 32.00 \space g \space O_2 } \space and \space \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 }$$
2. Use them as conversion factors to calculate the theoretical yield:
$$ 2.00 \times 10^{2} \space g \space C_3H_5N_3O_9 \times \frac{1 \space mole \space C_3H_5N_3O_9 }{ 227.10 \space g \space C_3H_5N_3O_9 } \times \frac{ 1 \space mole \space O_2 }{ 4 \space moles \space C_3H_5N_3O_9 } \times \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 } = 7.05 \space g \space O_2 $$
3. Calculation of percent yield: $$\frac{ 6.55 \space g \space O_2 }{ 7.05 \space g \space O_2 } \times 100\% = 92.9 \%$$
