Copper `(Cu)` and `(Zn)` react differently with `HNO_(3)` as follows:
`Cu + 4H^(o+) (aq) + 2NO_(3)^(ɵ) (aq) rarr 2NO_(2)(g) _ Cu^(2+) + 2H_(2)O`
`4Zn +10 H^(o+) (aq) + 2NO_(3)^(ɵ) (aq) rarr NH_(4)^(o+) + 4Zn^(2+) + 3H_(2)O`
What volume of `2.0 M HNO_(3)` would react with `10.0g` of a brass `(90.% Cu, 10.0% Zn)` according to the above equation?

Options

A. `~~ 100 mL`
B. `~~ 150 mL`
C. `~~ 200 mL`
D. `~~ 300 mL`

Answer

D

Explanation

Weight of `Cu = (10 xx 90)/(100) = 9.0 g`
Weight of `Zn = (10 xx 10)/(100) = 1.0 g`
Mol of `HNO_(3)` with `Cu = (9.0 g Cu)`
`((1 "mol" Cu)/(63.5 g))((4 "mol"HNO_(3))/(1 "mol"Cu))`
`= (9.0 xx 1 xx 4)/(63.5) = 0.56 "mol" HNO_(3)`
Mol of `HNO_(3)` with `Zn = (1.0 g)`
`((1 "mol"Zn)/(65.37))((10"mol" HNO_(3))/(4 "mol" Zn))`
`= (1.0 xx 10)/(65.37 xx 4)`
`= 0.038 "mol" HNO_(3)`
Total moles of `HNO_(3) = 0.56 + 0.38 = 0.598 ~~ 0.6`
`M xx V_(L) = "moles"`
`2.0 xx V_(L) = 0.6`
`V_(L) = 0.3 = 300 mL`