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 `NO_(2)` gas at `27^(@)C` and 1.0 atm pressure would be produced? if given 2 M `HNO_(3)` react with 10 g of a brass ( 90 % Cu 10% Zn) according to the above equation.

Options

A. `6.97 L`
B. `5.97 L`
C. `4.97 L`
D. `3.97 L`

Answer

A

Explanation

`NO_(2)` gas is produced only with `Cu` and `HNO_(3)`
Mol of `NO_(2)` with `Cu` and `HNO_(3)`
`= (9.0 g Cu) ((1 "mol" Cu)/(63.5 g))((2 "mol" NO_(2))/(1 "mol" Cu))`
`= (9 xx 1 xx 2)/(63.5 xx 1) = 0.283 "mol" NO_(2)`
`V = (nRT)/(P )`
`= ((0.283 "mol") (0.082 L atm mol^(-1) K^(-1)) xx 300 K)/(1.00 atm)`
`= 6.97 L`