Answer
The percent yield of carbon monoxide for this reaction is equal to 60.5 %
Work Step by Step
$ C $ : 12.01 g/mol
$$ \frac{1 \space mole \space C }{ 12.01 \space g \space C } \space and \space \frac{ 12.01 \space g \space C }{1 \space mole \space C }$$
$ CO $ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 1 )= 28.01 g/mol
$$ \frac{1 \space mole \space CO }{ 28.01 \space g \space CO } \space and \space \frac{ 28.01 \space g \space CO }{1 \space mole \space CO }$$
$$ 30.0 \space g \space C \times \frac{1 \space mole \space C }{ 12.01 \space g \space C } \times \frac{ 2 \space moles \space CO }{ 3 \space moles \space C } \times \frac{ 28.01 \space g \space CO }{1 \space mole \space CO } = 46.6 \space g \space CO $$
Calculation of percent yield: $$\frac{ 28.2 \space g \space CO }{ 46.6 \space g \space CO } \times 100\% = 60.5 \%$$
