In the Ostwald process, 4NH3 + 5O2 -> 4NO + 6H2O over Pt-Rh is followed by 2NO + O2 -> 2NO2 and 3NO2 + H2O -> 2HNO3 + NO with the NO recycled. If 17 g of NH3 gives nitric acid with an overall yield of 80%, the mass of HNO3 (molar mass 63 g mol^-1, N = 14, H = 1) obtained is

JEE Advanced levelSingle correct optionDifficulty 3 of 3df046

In the Ostwald process, $4 NH X_{3} + 5 O X_{2} 4 NO + 6 H X_{2} O$ over Pt-Rh is followed by $2 NO + O X_{2} 2 NO X_{2}$ and $3 NO X_{2} + H X_{2} O 2 HNO X_{3} + NO$ with the NO recycled. If $17 g$ of $NH X_{3}$ gives nitric acid with an overall yield of $80%$ , the mass of $HNO X_{3}$ (molar mass $63 g mo l^{- 1}$ , N = 14, H = 1) obtained is

With NO recycled the overall reaction is

NH X_{3} + 2 O X_{2} HNO X_{3} + H X_{2} O

(N 1, H 3, O 4 on both sides), so 1 mol

NH X_{3}

(17 g) gives 1 mol

HNO X_{3}

= 63 g at 100%. At 80% the mass is

63 \times 0.80 = 50.4 g