Nitrogen dioxide (4.6 g, NO2 = 46 g/mol) is passed into water in the presence of excess oxygen according to 4NO2 + O2 + 2H2O -> 4HNO3 (take HNO3 = 63 g/mol). The mass of nitric acid formed is

JEE Advanced levelSingle correct optionDifficulty 3 of 3pn183

Nitrogen dioxide (4.6 g, $NO X_{2}$ = 46 g/mol) is passed into water in the presence of excess oxygen according to $4 NO X_{2} + O X_{2} + 2 H X_{2} O 4 HNO X_{3}$ (take $HNO X_{3}$ = 63 g/mol). The mass of nitric acid formed is

Moles of

NO X_{2}

= 4.6/46 = 0.10. The equation gives 1 mol

HNO X_{3}

per mol

NO X_{2}

, so 0.10 mol

HNO X_{3}

forms. Mass =

0.10 \times 63 = 6.30