Answer
a. 77.5 g of $Na_2O$ are produced when 57.5 g of $Na$ reacts.
b. 6.26 g of $O_2$ are needed to react with 18.0 g of $Na$
c. 19.4 g of $O_2$ are needed to produce 75.0 g of $Na_2O$
Work Step by Step
- Identify the conversion factors and use them to calculate the final mass.
a.
$$ \frac{ 4 \space moles \space Na }{ 2 \space moles \space Na_2O } \space and \space \frac{ 2 \space moles \space Na_2O }{ 4 \space moles \space Na }$$
$ Na $ : 22.99 g/mol
$$ \frac{1 \space mole \space Na }{ 22.99 \space g \space Na } \space and \space \frac{ 22.99 \space g \space Na }{1 \space mole \space Na }$$
$ Na_2O $ : ( 22.99 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 61.98 g/mol
$$ \frac{1 \space mole \space Na_2O }{ 61.98 \space g \space Na_2O } \space and \space \frac{ 61.98 \space g \space Na_2O }{1 \space mole \space Na_2O }$$
$$ 57.5 \space g \space Na \times \frac{1 \space mole \space Na }{ 22.99 \space g \space Na } \times \frac{ 2 \space moles \space Na_2O }{ 4 \space moles \space Na } \times \frac{ 61.98 \space g \space Na_2O }{1 \space mole \space Na_2O } = 77.5 \space g \space Na_2O $$
b. $$ \frac{ 4 \space moles \space Na }{ 1 \space mole \space O_2 } \space and \space \frac{ 1 \space mole \space O_2 }{ 4 \space moles \space Na }$$
$ Na $ : 22.99 g/mol
$$ \frac{1 \space mole \space Na }{ 22.99 \space g \space Na } \space and \space \frac{ 22.99 \space g \space Na }{1 \space mole \space Na }$$
$ O_2 $ : ( 16.00 $\times$ 2 )= 32.00 g/mol
$$ \frac{1 \space mole \space O_2 }{ 32.00 \space g \space O_2 } \space and \space \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 }$$
$$ 18.0 \space g \space Na \times \frac{1 \space mole \space Na }{ 22.99 \space g \space Na } \times \frac{ 1 \space mole \space O_2 }{ 4 \space moles \space Na } \times \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 } = 6.26 \space g \space O_2 $$
c.
$$ \frac{ 2 \space moles \space Na_2O }{ 1 \space mole \space O_2 } \space and \space \frac{ 1 \space mole \space O_2 }{ 2 \space moles \space Na_2O }$$
$ Na_2O $ : 61.98 g/mol
$$ \frac{1 \space mole \space Na_2O }{ 61.98 \space g \space Na_2O } \space and \space \frac{ 61.98 \space g \space Na_2O }{1 \space mole \space Na_2O }$$
$ O_2 $ : 32.00 g/mol
$$ \frac{1 \space mole \space O_2 }{ 32.00 \space g \space O_2 } \space and \space \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 }$$
$$ 75.0 \space g \space Na_2O \times \frac{1 \space mole \space Na_2O }{ 61.98 \space g \space Na_2O } \times \frac{ 1 \space mole \space O_2 }{ 2 \space moles \space Na_2O } \times \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 } = 19.4 \space g \space O_2 $$
