`Mg` can reduce `NO_(3)^(-)` to `NH_(3)` in basic solution:
`NO_(3)^(-) + Mg_((s)) + H_(2)O rarr Mg (OH)_(2(s)) + OH_((aq.))^(-) + NH_(3(g))`
A `25.0 mL` sample of `NO_(3)^(-)` solution was treated with `Mg`. The `NH_(3(g))` was passed into `50mL` to `0.15N HCl`. The excess `HCl` required `32.10 mL` of `0.10M NaOH` for its neutralisation. What was the molarity of `NO_(3)^(-)` ions in the original sample?

Answer

`0.1716`