Answer
81.0 g of $CO_2$
Work Step by Step
$ C_3H_8 $ : ( 1.008 $\times$ 8 )+ ( 12.01 $\times$ 3 )= 44.09 g/mol
$$ \frac{1 \space mole \space C_3H_8 }{ 44.09 \space g \space C_3H_8 } \space and \space \frac{ 44.09 \space g \space C_3H_8 }{1 \space mole \space C_3H_8 }$$
$ CO_2 $ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 44.01 g/mol
$$ \frac{1 \space mole \space CO_2 }{ 44.01 \space g \space CO_2 } \space and \space \frac{ 44.01 \space g \space CO_2 }{1 \space mole \space CO_2 }$$
$$ 45.0 \space g \space C_3H_8 \times \frac{1 \space mole \space C_3H_8 }{ 44.09 \space g \space C_3H_8 } \times \frac{ 3 \space moles \space CO_2 }{ 1 \space mole \space C_3H_8 } \times \frac{ 44.01 \space g \space CO_2 }{1 \space mole \space CO_2 } = 135 \space g \space CO_2 $$
$$actual \space yield =\frac{ percent \space yield \times theoretical \space yield }{100\%}$$$$actual \space yield =\frac{( 60.0 \%)\times ( 135 \space g \space CO_2 )}{100\%} = 81.0 \space g \space CO_2 $$
