Answer
a. 579 kJ are released when 125 g of $Cl_2$ reacts with silicon.
b. 89 kJ are absorbed when 278 g of $PCl_5$ reacts.
Work Step by Step
a.
- Find the conversion factors:
$ Cl_2 $ : ( 35.45 $\times$ 2 )= 70.90 g/mol
$$ \frac{1 \space mole \space Cl_2 }{ 70.90 \space g \space Cl_2 } \space and \space \frac{ 70.90 \space g \space Cl_2 }{1 \space mole \space Cl_2 }$$
According to the balanced equation:
$$ \frac{ 2 \space moles \space Cl_2 }{ -657 \space kJ} \space and \space \frac{ -657 \space kJ}{ 2 \space moles \space Cl_2 }$$
- Find the total energy using these conversion factors:
$$ 125 \space g \space Cl_2 \times \frac{1 \space mole \space Cl_2 }{ 70.90 \space g \space Cl_2 } \times \frac{ -657 \space kJ}{ 2 \space moles \space Cl_2 } = -579 \space kJ$$
Since $\Delta H$ is negative, 579 kJ were released in this reaction.
b.
- Find the conversion factors:
$ PCl_5 $ : ( 35.45 $\times$ 5 )+ ( 30.974 $\times$ 1 )= 208.22 g/mol
$$ \frac{1 \space mole \space PCl_5 }{ 208.22 \space g \space PCl_5 } \space and \space \frac{ 208.22 \space g \space PCl_5 }{1 \space mole \space PCl_5 }$$
According to the balanced equation:
$$ \frac{ 1 \space mole \space PCl_5 }{ 67 \space kJ} \space and \space \frac{ 67 \space kJ}{ 1 \space mole \space PCl_5 }$$
- Find the total energy using these conversion factors:
$$ 278 \space g \space PCl_5 \times \frac{1 \space mole \space PCl_5 }{ 208.22 \space g \space PCl_5 } \times \frac{ 67 \space kJ}{ 1 \space mole \space PCl_5 } = 89 \space kJ$$
Since $\Delta H$ is positive, 89 kJ were absorbed in this reaction.
