When copper is treated with a certain concentration of nitric acid, nitric oxide and nitrogen dioxide are liberated in equal volumes according to the equation
`xCu+yHNO_(3)rarrCu(NO_(3))_(2)+NO+NO_(2)+H_(2)O`
The coefficients `x` and `y` are

Options

A. 2 and 3
B. 2 and 6
C. 1 and 3
D. 3 and 8

Answer

B

Explanation

Copper reacts with conc. `HNO_(3)` as follows :
`Cu+4HNO_(3)rarrCu(NO_(3))_(2)+2H_(2)O+2NO_(2)`
Copper reacts with dil. `HNO_(3)` as follows :
`3Cu+8HNO_(3)rarr3Cu(NO_(3))_(2)+4H_(2)O+2NO`
As equal volumes of `NO_(2)` and NO are evolved, add the two equations to get another equation
`4Cu+12HNO_(3)rarr4Cu(NO_(3))_(2)+6H_(2)O+2NO_(2)+2NO`
or `4Cu+12HNO_(3)rarr4Cu(NO_(3))_(2)+6H_(2)O+2NO_(2)+2NO`.