In the reaction between dilute nitric acid and magnesium metal, (a) how many moles of `HNO_(3)` reacts with a gram atom of metal? and (b) how many moles of `HNO_(3)` is oxidised by a gram atom of metal? 

Explanation

The skeleton equation is
`Mg+HNO_(3) to Mg(NO_(3))_(2)+NH_(4)NO_(3)+H_(2)O`
The balanced equation is
`4Mg +10HNO_(3) to 4Mg(NO_(3))_(2)+NH_(4)NO_(3)+3H_(2)O`
(a) Number of moles of `HNO_(3)` that reacts with a gram atom of Mg=10/4=2.5
(b) `NO_(3)^(-)` is a spectator ion. It is neglected in the equation.
`4Mg+NO_(3)^(-)+10H^(+) to 4Mg^(2+)+NH_(4)^(+)+3H_(O)`
Number of moles of `HNO_(3)` that is reduced by a gram atom of Mg=0.25.