Answer
a. Endothermic
b. 12.3 g of glucose $(C_6H_{12}O_6)$
c. 390 kJ is needed to produce 25.0 g of glucose.
Work Step by Step
a. Since the energy is in the reactants side of the reaction, it is an endothermic reaction.
b.
$ CO_2 $ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 44.01 g/mol
$$ \frac{1 \space mole \space CO_2 }{ 44.01 \space g \space CO_2 } \space and \space \frac{ 44.01 \space g \space CO_2 }{1 \space mole \space CO_2 }$$
$ C_6H_{12}O_6 $ : ( 1.008 $\times$ 12 )+ ( 12.01 $\times$ 6 )+ ( 16.00 $\times$ 6 )= 180.16 g/mol
$$ \frac{1 \space mole \space C_6H_{12}O_6 }{ 180.16 \space g \space C_6H_{12}O_6 } \space and \space \frac{ 180.16 \space g \space C_6H_{12}O_6 }{1 \space mole \space C_6H_{12}O_6 }$$
$$ 18.0 \space g \space CO_2 \times \frac{1 \space mole \space CO_2 }{ 44.01 \space g \space CO_2 } \times \frac{ 1 \space mole \space C_6H_{12}O_6 }{ 6 \space moles \space CO_2 } \times \frac{ 180.16 \space g \space C_6H_{12}O_6 }{1 \space mole \space C_6H_{12}O_6 } = 12.3 \space g \space C_6H_{12}O_6 $$
c.
- Find the conversion factors:
$ C_6H_{12}O_6 $ : ( 1.008 $\times$ 12 )+ ( 12.01 $\times$ 6 )+ ( 16.00 $\times$ 6 )= 180.16 g/mol
$$ \frac{1 \space mole \space C_6H_{12}O_6 }{ 180.16 \space g \space C_6H_{12}O_6 } \space and \space \frac{ 180.16 \space g \space C_6H_{12}O_6 }{1 \space mole \space C_6H_{12}O_6 }$$
According to the balanced equation:
$$ \frac{ 1 \space mole \space C_6H_{12}O_6 }{ 680 \space kcal} \space and \space \frac{ 680 \space kcal}{ 1 \space mole \space C_6H_{12}O_6 }$$
And finally, since the question requires a result in kJ:
$$\frac{4.184 \space kJ}{1 \space kcal} and \frac{1 \space kcal}{4.184 \space kJ}$$
- Find the total energy using these conversion factors:
$$ 25.0 \space g \space C_6H_{12}O_6 \times \frac{1 \space mole \space C_6H_{12}O_6 }{ 180.16 \space g \space C_6H_{12}O_6 } \times \frac{ 680 \space kcal}{ 1 \space mole \space C_6H_{12}O_6 } \times \frac{4.184 \space kJ}{1 \space kcal} = 390 \space kJ$$
