Answer
a. The percent yield of carbon disulfide in this case is equal to 71.0 %
b. a. The percent yield of carbon disulfide in this case is equal to 63.2 %
Work Step by Step
a.
$ C $ : 12.01 g/mol
$$ \frac{1 \space mole \space C }{ 12.01 \space g \space C } \space and \space \frac{ 12.01 \space g \space C }{1 \space mole \space C }$$
$ CS_2 $ : ( 12.01 $\times$ 1 )+ ( 32.06 $\times$ 2 )= 76.13 g/mol
$$ \frac{1 \space mole \space CS_2 }{ 76.13 \space g \space CS_2 } \space and \space \frac{ 76.13 \space g \space CS_2 }{1 \space mole \space CS_2 }$$
$$ 40.0 \space g \space C \times \frac{1 \space mole \space C }{ 12.01 \space g \space C } \times \frac{ 1 \space mole \space CS_2 }{ 5 \space moles \space C } \times \frac{ 76.13 \space g \space CS_2 }{1 \space mole \space CS_2 } = 50.7 \space g \space CS_2 $$
Calculation of percent yield: $$\frac{ 36.0 \space g \space CS_2 }{ 50.7 \space g \space CS_2 } \times 100\% = 71.0 \%$$
b.
$ SO_2 $ : ( 16.00 $\times$ 2 )+ ( 32.06 $\times$ 1 )= 64.06 g/mol
$$ \frac{1 \space mole \space SO_2 }{ 64.06 \space g \space SO_2 } \space and \space \frac{ 64.06 \space g \space SO_2 }{1 \space mole \space SO_2 }$$
$ CS_2 $ : ( 12.01 $\times$ 1 )+ ( 32.06 $\times$ 2 )= 76.13 g/mol
$$ \frac{1 \space mole \space CS_2 }{ 76.13 \space g \space CS_2 } \space and \space \frac{ 76.13 \space g \space CS_2 }{1 \space mole \space CS_2 }$$
$$ 32.0 \space g \space SO_2 \times \frac{1 \space mole \space SO_2 }{ 64.06 \space g \space SO_2 } \times \frac{ 1 \space mole \space CS_2 }{ 2 \space moles \space SO_2 } \times \frac{ 76.13 \space g \space CS_2 }{1 \space mole \space CS_2 } = 19.0 \space g \space CS_2 $$
Calculation of percent yield: $$\frac{ 12.0 \space g \space CS_2 }{ 19.0 \space g \space CS_2 } \times 100\% = 63.2 \%$$
