Answer
(a) Not balanced.
The balanced equation is: $2Al + Fe_2O_3 -- \gt Al_2O_3 + 2Fe$
(b) It is already balanced.
(c) Not balanced.
$4Au + 8NaCN + O_2 + 2H_2O --\gt 4NaAu(CN)_2 + 4NaOH$
Work Step by Step
(a) $Al + Fe_2O_3 -- \gt Al_2O_3 + Fe$
The $Al$ compounds are not balanced: put a 2 on the one on the left side.
$2Al + Fe_2O_3 -- \gt Al_2O_3 + Fe$
Now, the only non balanced element is $Fe$. We can adjust that by putting a 2 on the $Fe$ on the right side.
$2Al + Fe_2O_3 -- \gt Al_2O_3 + 2Fe$
(b) The number of carbons is equal to 6 on both sides, that for hydrogen is 12 on both sides, and we have 18 oxygens on both sides.
(c)
Balance the $CN$ compounds.
$Au + 2NaCN + O_2 + H_2O --\gt NaAu(CN)_2 + NaOH$
Balance the hydrogen number:
$Au + 2NaCN + O_2 + H_2O --\gt NaAu(CN)_2 + 2NaOH$
We need to add 1 $Na$ to the left side, but the $CN$ number has to still the same, we can do that by doing this:
$Au + 4NaCN + O_2 + H_2O --\gt 2NaAu(CN)_2 + 2NaOH$
Balance $Au:$
$2Au + 4NaCN + O_2 + H_2O --\gt 2NaAu(CN)_2 + 2NaOH$
Now, the only unbalanced element is oxygen; trying to balance it:
$Au + 4NaCN + \frac{1}{2}O_2 + H_2O --\gt 2NaAu(CN)_2 + 2NaOH$
Multiply all coefficients by 2.
$4Au + 8NaCN + O_2 + 2H_2O --\gt 4NaAu(CN)_2 + 4NaOH$
