Calculate the number of moles of `Cu` and `HNO_(3)` to give `NO` and `NO_(2)` in the `(2:1)` molar ratio.

Explanation

`{:(CutoCu^(2+)+2e^(Theta)xx3),(4H^(Theta)+3e^(Theta)+NO_(3)^(Theta)toNO+2H_(2)Oxx2),(ulbar(3Cu+8H^(o+)+2NO_(3)^(Theta)toCu^(2+)+2NO_(2)+2H_(2)O)):}`
or

or
`Cu + 4HNO_(3)(conc) rarr Cu(NO_(3))_(2) + 2NO_(2) + 4H_(2)O` …(ii)
To get the `NO` and `NO_(2)` in the molar ratio of `2:1`, multiply equation (i) by `2` and then add equations (i) and (ii), to get `7Cu + 20HNO_(3) rarr 7Cu(NO_(3))_(2) + 4NO + 4NO_(2) + 10H_(2)O`
Hence, number of moles of `Cu = 7`
Number of moles of `HNO_(3) = 20`