Answer
a. $$4Al(s) + 3O_2(g) \longrightarrow 2Al_2O_3$$
b. This is a combination reaction.
c. 3.38 moles of $O_2$ are needed to react with 4.50 moles of $Al$
d. 94.9 g of $Al_2O_3$ are produced.
e. 17.0 g of $Al_2O_3$.
Work Step by Step
a. $Al(s) + O_2(g) \longrightarrow Al_2O_3$
- Balance the number of aluminum atoms:
$$2Al(s) + O_2(g) \longrightarrow Al_2O_3$$
- Balance the number of oxygens:
$$2Al(s) + \frac{3}{2}O_2(g) \longrightarrow Al_2O_3$$
- In order to remove the fraction, multiply all coeficients by 2:
$$4Al(s) + 3O_2(g) \longrightarrow 2Al_2O_3$$
b. Since this equation follows the pattern: $A + B \longrightarrow AB$, it is a combination reaction.
c. $$ 4.50 \space moles \space Al \times \frac{ 3 \space moles \ O_2 }{ 4 \space moles \space Al } = 3.38 \space moles \space O_2 $$
d.
$ Al $ : 26.98 g/mol
$$ \frac{1 \space mole \space Al }{ 26.98 \space g \space Al } \space and \space \frac{ 26.98 \space g \space Al }{1 \space mole \space Al }$$
$ Al_2O_3 $ : ( 26.98 $\times$ 2 )+ ( 16.00 $\times$ 3 )= 101.96 g/mol
$$ \frac{1 \space mole \space Al_2O_3 }{ 101.96 \space g \space Al_2O_3 } \space and \space \frac{ 101.96 \space g \space Al_2O_3 }{1 \space mole \space Al_2O_3 }$$
$$ 50.2 \space g \space Al \times \frac{1 \space mole \space Al }{ 26.98 \space g \space Al } \times \frac{ 2 \space moles \space Al_2O_3 }{ 4 \space moles \space Al } \times \frac{ 101.96 \space g \space Al_2O_3 }{1 \space mole \space Al_2O_3 } = 94.9 \space g \space Al_2O_3 $$
e.
$ O_2 $ : ( 16.00 $\times$ 2 )= 32.00 g/mol
$$ \frac{1 \space mole \space O_2 }{ 32.00 \space g \space O_2 } \space and \space \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 }$$
$ Al_2O_3 $ : 101.96 g/mol
$$ \frac{1 \space mole \space Al_2O_3 }{ 101.96 \space g \space Al_2O_3 } \space and \space \frac{ 101.96 \space g \space Al_2O_3 }{1 \space mole \space Al_2O_3 }$$
$$ 8.00 \space g \space O_2 \times \frac{1 \space mole \space O_2 }{ 32.00 \space g \space O_2 } \times \frac{ 2 \space moles \space Al_2O_3 }{ 3 \space moles \space O_2 } \times \frac{ 101.96 \space g \space Al_2O_3 }{1 \space mole \space Al_2O_3 } = 17.0 \space g \space Al_2O_3 $$
