Answer
$a) Al(ClO_{3})_{3}$
$b) (NH_{4})_{2}O$
$c) Mg(HCO_{3})_{2}$
$d) NaNO_{2}$
$e) Cu_{2}SO_{4}$
Work Step by Step
$a) Al^{3+} ClO_{3}^{-} = Al(ClO_{3})_{3}$
$b) NH_{4}^{+} O^{2-} = (NH_{4})_{2}O$
$c) Mg^{2+} HCO_{3}^{-} = Mg(HCO_{3})_{2}$
$d) Na^{+} NO_{2}^{-} = NaNO_{2}$
$e) Cu^{+} SO_{4}^{2-} = Cu_{2}SO_{4}$
