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

Explanation

The balanced equation is
`Cu to Cu^(2+)+2e^(-) ...........(i)`
`H^(+)+e^(-)+HNO_(3) to NO_(2)+H_(2)O ...............(ii) `
`3H^(+)+ 3e^(-) +HNO_(3) to NO+2H_(2)O ......(iii)`
Multiplying (i) by 2 and adding (i), (ii) and (iii), we get
`2Cu+4H^(+) +2HNO_(3) to 2Cu^(2+)+NO_(2)`
`+NO+3H_(2)O or 2Cu+6HNO_(3) to 2Cu(NO_(3))_(2)`
`to NO_(2)+NO+3H_(2)O`
x and y are 2 and 6