Answer
a. The theoretical yield is equal to 74.4 g of $CO_2$
b. The percent yield is equal to 86.0 %
Work Step by Step
a.
$ C_2H_2 $ : ( 1.008 $\times$ 2 )+ ( 12.01 $\times$ 2 )= 26.04 g/mol
$$ \frac{1 \space mole \space C_2H_2 }{ 26.04 \space g \space C_2H_2 } \space and \space \frac{ 26.04 \space g \space C_2H_2 }{1 \space mole \space C_2H_2 }$$
$ CO_2 $ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 44.01 g/mol
$$ \frac{1 \space mole \space CO_2 }{ 44.01 \space g \space CO_2 } \space and \space \frac{ 44.01 \space g \space CO_2 }{1 \space mole \space CO_2 }$$
$$ 22.0 \space g \space C_2H_2 \times \frac{1 \space mole \space C_2H_2 }{ 26.04 \space g \space C_2H_2 } \times \frac{ 4 \space moles \space CO_2 }{ 2 \space moles \space C_2H_2 } \times \frac{ 44.01 \space g \space CO_2 }{1 \space mole \space CO_2 } = 74.4 \space g \space CO_2 $$
b.
Calculation of percent yield: $$\frac{ 64.0 \space g \space CO_2 }{ 74.4 \space g \space CO_2 } \times 100\% = 86.0 \%$$
