Answer
The percent yield of $C_2H_6$ is equal to 75.9 %
Work Step by Step
- Find the theoretical yield:
$ C_2H_2 $ : ( 1.008 $\times$ 2 )+ ( 12.01 $\times$ 2 )= 26.04 g/mol
$$ \frac{1 \space mole \space C_2H_2 }{ 26.04 \space g \space C_2H_2 } \space and \space \frac{ 26.04 \space g \space C_2H_2 }{1 \space mole \space C_2H_2 }$$
$ C_2H_6 $ : ( 1.008 $\times$ 6 )+ ( 12.01 $\times$ 2 )= 30.07 g/mol
$$ \frac{1 \space mole \space C_2H_6 }{ 30.07 \space g \space C_2H_6 } \space and \space \frac{ 30.07 \space g \space C_2H_6 }{1 \space mole \space C_2H_6 }$$
$$ 28.0 \space g \space C_2H_2 \times \frac{1 \space mole \space C_2H_2 }{ 26.04 \space g \space C_2H_2 } \times \frac{ 1 \space mole \space C_2H_6 }{ 1 \space mole \space C_2H_2 } \times \frac{ 30.07 \space g \space C_2H_6 }{1 \space mole \space C_2H_6 } = 32.3 \space g \space C_2H_6 $$
Calculation of percent yield: $$\frac{ 24.5 \space g \space C_2H_6 }{ 32.3 \space g \space C_2H_6 } \times 100\% = 75.9 \%$$
