Answer
a. Exothermic.
b. 35.5 kcal
c. 176 g of $C_2H_6O$
Work Step by Step
a. Since the energy is in the products, this is an exothermic reaction.
b.
- Find the conversion factors:
$ C_2H_6O $ : ( 1.008 $\times$ 6 )+ ( 12.01 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 46.07 g/mol
$$ \frac{1 \space mole \space C_2H_6O }{ 46.07 \space g \space C_2H_6O } \space and \space \frac{ 46.07 \space g \space C_2H_6O }{1 \space mole \space C_2H_6O }$$
According to the balanced equation:
$$ \frac{ 1 \space mole \space C_2H_6O }{ 327 \space kcal} \space and \space \frac{ 327 \space kcal}{ 1 \space mole \space C_2H_6O }$$
- Find the total energy using these conversion factors:
$$ 5.00 \space g \space C_2H_6O \times \frac{1 \space mole \space C_2H_6O }{ 46.07 \space g \space C_2H_6O } \times \frac{ 327 \space kcal}{ 1 \space mole \space C_2H_6O } = 35.5 \space kcal$$
c.- Find the conversion factors:
$ C_2H_6O $ : ( 1.008 $\times$ 6 )+ ( 12.01 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 46.07 g/mol
$$ \frac{1 \space mole \space C_2H_6O }{ 46.07 \space g \space C_2H_6O } \space and \space \frac{ 46.07 \space g \space C_2H_6O }{1 \space mole \space C_2H_6O }$$
According to the balanced equation:
$$ \frac{ 1 \space mole \space C_2H_6O }{ 327 \space kcal} \space and \space \frac{ 327 \space kcal}{ 1 \space mole \space C_2H_6O }$$
And finally, since the energy is 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:
$$ 1250 \space kJ \times \frac{1 \space kcal}{4.184 \space kJ} \times \frac{ 1 \space mole \space C_2H_6O }{ 327 \space kcal} \times \frac{ 46.07 \space g \space C_2H_6O }{1 \space moles \space C_2H_6O } = 176 \space g \space C_2H_6O$$
