One mole of tin is treated with concentrated HNO3 (Sn + 4HNO3 -> SnO2.H2O + 4NO2 + H2O) and one mole of lead is treated separately with dilute HNO3 (3Pb + 8HNO3 -> 3Pb(NO3)2 + 2NO + 4H2O). The total number of moles of nitrogen oxides (NO2 plus NO) evolved is

JEE Advanced levelSingle correct optionDifficulty 3 of 3pb108

One mole of tin is treated with concentrated $HNO X_{3}$ ( $Sn + 4 HNO X_{3} SnO X_{2} \cdot H X_{2} O + 4 NO X_{2} + H X_{2} O$ ) and one mole of lead is treated separately with dilute $HNO X_{3}$ ( $3 Pb + 8 HNO X_{3} 3 Pb (NO X_{3}) X_{2} + 2 NO + 4 H X_{2} O$ ). The total number of moles of nitrogen oxides ( $NO X_{2}$ plus $NO$ ) evolved is

Tin gives 4 mol

NO X_{2}

per mole (Sn goes from 0 to

+ 4

, four electrons). For lead, 3 mol Pb give 2 mol

NO

, so 1 mol Pb gives

2/3 = 0.67

mol

NO

(Pb goes to

+ 2

, two electrons, each

NO

takes three). Total

= 4 + 0.67 = 4.67

mol.