2 g of brass containing Cu and Zn only reacts with 3M `HNO_(3)` solution. Following are the reactions taking place `Cu(s) + HNO_(3) (aq) to Cu^(2+) (aq) +NO_(2)(g) + H_(2)O(I)`
`Zn(s) + H^(+)(aq) + NO_(3)^(-)(aq) to NH_(4)^(+) + Zn^(2+)(aq) + H_(2)O(l)`
The liberated `NO_(2)(g)` was found to be 1.04 L at `25^(@)`C and 1 atm `[Cu=63.5 , Zn=65.4]`
The percentage by mass of Cu in brass was

Explanation

`CutoCu^(2+)+2e^(-)`
`undersetunderset(x=5)(1+x-6=0)[H^(+)+e^(-)+HNO_(3))toundersetunderset(x=4)(x-4=0)(NO_(2))+H_(2)O`
`Cu+2H^(o+)+2HCO_(3)to2NO_(2)2H_(2)O+Cu^(2+)`
`Cu+4HNO_(3)to2NO_(2)+2H_(2)O+Cu^(2+)+2NO_(3)^(ɵ)`
or `Cu+4HNO_(3)to2NO_(2)+2H_(2)O+Cu^(2+)+2H_(2)O`
1 " mol of "`Cu=4 " mol of "HNO_(3)-=2 " mol of "NO_(2)`
Mole of `NO_(2)=(PV)/(RT)=(1xx1xx0.4)/(0.082xx298)=0.0425`
mole of `Cu=(0.0425)/(2)=(0.0425)/(2)xx63.5=1.3462g`
Weight of `Zn=1.5-1.3462=0.1538g`
`% Zn=10.25%`
`% of Cu=89.75%`
`[ZntoZn^(+2)+2e^(-)]xx4`
`8e^(-)+undersetunderset(x=5)(x-6=-1)(NO_(3)^(ɵ))+10H^(o+)toundersetunderset(x=-3)(x+4=+1)(NH_(4)^(o+)+3H_(2)O)`
or `4An+10HNO_(3)toNH_(4)^(o+)+4An^(2+)+3H_(2)O`
`4Zn+10HNO_(3)toNH_(4)NO_(3)+4Zn(NO_(3))_(2)+3H_(2)O`
`=3xxV`
`V=0.2011 L =20.11mL`