What volume of `3.0 MHNO_(3)` can reacat completely with `15.0 g` brass `(90% Cu and 10%Zn)` according to equation:
`Cu+4H_((aq))^(+)+2NO_(3(aq))^(-)rarr2NO_(2(g))+Cu^(2+)+2H_(2)O`
`4Zn+10H_((aq))^(+)+NO_(3(aq))^(-)rarrNH_(4)^(+)+4Zn^(2+)+3H_(2)O`
Also report what volume of `NO_(2)` gas at `25^(@)C` and `1.0 atm` will be prodeced?

Explanation

`Cu+4HNO_(3)rarr2NO_(2)+Cu(NO_(3))_(2)+2H_(2)O`
`4Zn+10HNO_(3)rarrZn(NO_(3))_(2)+NH_(4)NO_(3)+3H_(2)O`
Mole of `HNO_(3)` used for `Cu=4xx"mole of"Cu`
`=4xx(15xx90)/(100xx63.6)`
`3xxV_(1)= 0.85`
`V_(1)= 0.283 "litre"=283 mL`
Mole of `HNO_(3)` used for `Zn=(10)/(4)xx"mole of" Zn`
`=(10)/(4)xx(15xx10)/(100xx65)`
`3xxV_(2)= 0.058`
`:. V_(2)= 0.019"litre"=19mL`
`:.` Total `HNO_(3)= 283+19= 302 mL`
Also mole of `NO_(2)=2xx"mole of"Cu`
`=(2xx15xx90)/(100xx63.6)=0.425`
`:. PV=nRT`
`V=(0.425xx0.0821xx298)/(1)= 10.38 "litre"`