Answer
a. 850 kJ.
b. 278 kJ.
Work Step by Step
a.
- Find the conversion factors:
$ CH_3OH $ : ( 1.008 $\times$ 4 )+ ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 1 )= 32.04 g/mol
$$ \frac{1 \space mole \space CH_3OH }{ 32.04 \space g \space CH_3OH } \space and \space \frac{ 32.04 \space g \space CH_3OH }{1 \space mole \space CH_3OH }$$
According to the balanced equation:
$$ \frac{ 2 \space moles \space CH_3OH }{ -726 \space kJ} \space and \space \frac{ -726 \space kJ}{ 2 \space moles \space CH_3OH }$$
- Find the total energy using these conversion factors:
$$ 75.0 \space g \space CH_3OH \times \frac{1 \space mole \space CH_3OH }{ 32.04 \space g \space CH_3OH } \times \frac{ -726 \space kJ}{ 2 \space moles \space CH_3OH } = -850. \space kJ$$
b.
- Find the conversion factors:
$ Ca(OH)_2 $ : ( 40.08 $\times$ 1 )+ ( 1.008 $\times$ 2 )+ ( 16.00 $\times$ 2 )= 74.10 g/mol
$$ \frac{1 \space mole \space Ca(OH)_2 }{ 74.10 \space g \space Ca(OH)_2 } \space and \space \frac{ 74.10 \space g \space Ca(OH)_2 }{1 \space mole \space Ca(OH)_2 }$$
According to the balanced equation:
$$ \frac{ 1 \space mole \space Ca(OH)_2 }{ 65.3 \space kJ} \space and \space \frac{ 65.3 \space kJ}{ 1 \space mole \space Ca(OH)_2 }$$
- Find the total energy using these conversion factors:
$$ 315 \space g \space Ca(OH)_2 \times \frac{1 \space mole \space Ca(OH)_2 }{ 74.10 \space g \space Ca(OH)_2 } \times \frac{ 65.3 \space kJ}{ 1 \space mole \space Ca(OH)_2 } = 278 \space kJ$$
