Answer
$$3.48 \times 10^3 \space g \space C_6H_{14} $$
Work Step by Step
1. Calculate the molar masses:
$ C_2H_4 $ : ( 1.008 $\times$ 4 )+ ( 12.01 $\times$ 2 )= 28.05 g/mol
$$ \frac{1 \space mole \space C_2H_4 }{ 28.05 \space g \space C_2H_4 } \space and \space \frac{ 28.05 \space g \space C_2H_4 }{1 \space mole \space C_2H_4 }$$
$ C_6H_{14} $ : ( 1.008 $\times$ 14 )+ ( 12.01 $\times$ 6 )= 86.17 g/mol
$$ \frac{1 \space mole \space C_6H_{14} }{ 86.17 \space g \space C_6H_{14} } \space and \space \frac{ 86.17 \space g \space C_6H_{14} }{1 \space mole \space C_6H_{14} }$$
2. Find the theoretical amount of reactant:
$$ 481 \space g \space C_2H_4 \times \frac{1 \space mole \space C_2H_4 }{ 28.05 \space g \space C_2H_4 } \times \frac{ 1 \space mole \space C_6H_{14} }{ 1 \space mole \space C_2H_4 } \times \frac{ 86.17 \space g \space C_6H_{14} }{1 \space mole \space C_6H_{14} } = 1480 \space g \space C_6H_{14} $$
$$ necessary \space reactant =\frac{ 100\% \times theoretical \space reactant }{ percent \space yield }$$$$ necessary \space reactant =\frac{( 100 \%)\times ( 1480 \space g \space C_6H_{14} )}{ 42.5 \%} = 3.48 \times 10^3 \space g \space C_6H_{14} $$
