Answer
a. 242 g of glucose are required to produce 124 g of ethanol.
b. 123 g of ethanol are formed from 0.240 kg of glucose.
Work Step by Step
a.
$ C_2H_6O $ : ( 1.008 $\times$ 6 )+ ( 12.01 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 46.07 g/mol
$$ \frac{1 \space mole \space C_2H_6O }{ 46.07 \space g \space C_2H_6O } \space and \space \frac{ 46.07 \space g \space C_2H_6O }{1 \space mole \space C_2H_6O }$$
$ C_6H_{12}O_6 $ : ( 1.008 $\times$ 12 )+ ( 12.01 $\times$ 6 )+ ( 16.00 $\times$ 6 )= 180.16 g/mol
$$ \frac{1 \space mole \space C_6H_{12}O_6 }{ 180.16 \space g \space C_6H_{12}O_6 } \space and \space \frac{ 180.16 \space g \space C_6H_{12}O_6 }{1 \space mole \space C_6H_{12}O_6 }$$
$$ 124 \space g \space C_2H_6O \times \frac{1 \space mole \space C_2H_6O }{ 46.07 \space g \space C_2H_6O } \times \frac{ 1 \space mole \space C_6H_{12}O_6 }{ 2 \space moles \space C_2H_6O } \times \frac{ 180.16 \space g \space C_6H_{12}O_6 }{1 \space mole \space C_6H_{12}O_6 } = 242 \space g \space C_6H_{12}O_6 $$
b.
$ C_6H_{12}O_6 $ : ( 1.008 $\times$ 12 )+ ( 12.01 $\times$ 6 )+ ( 16.00 $\times$ 6 )= 180.16 g/mol
$$ \frac{1 \space mole \space C_6H_{12}O_6 }{ 180.16 \space g \space C_6H_{12}O_6 } \space and \space \frac{ 180.16 \space g \space C_6H_{12}O_6 }{1 \space mole \space C_6H_{12}O_6 }$$
$ C_2H_6O $ : ( 1.008 $\times$ 6 )+ ( 12.01 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 46.07 g/mol
$$ \frac{1 \space mole \space C_2H_6O }{ 46.07 \space g \space C_2H_6O } \space and \space \frac{ 46.07 \space g \space C_2H_6O }{1 \space mole \space C_2H_6O }$$
$$1 \space kg = 1000 \space g$$
$$ 0.240 \space kg \space C_6H_{12}O_6 \times \frac{1000 \space g}{1 \space kg} \times \frac{1 \space mole \space C_6H_{12}O_6 }{ 180.16 \space g \space C_6H_{12}O_6 } \times \frac{ 2 \space moles \space C_2H_6O }{ 1 \space mole \space C_6H_{12}O_6 } \times \frac{ 46.07 \space g \space C_2H_6O }{1 \space mole \space C_2H_6O } = 123 \space g \space C_2H_6O $$
