200 ml , solution of HNO 3 having 63% (wt. by volume) is taken and 10 gm of impure zinc sample mixed with this solution. Now reaction mixture is heated so that NO 2 gas is removed and resultant so

Question

200 ml , solution of HNO 3 ​ having 63% (wt. by volume) is taken and 10 gm of impure zinc sample mixed with this solution. Now reaction mixture is heated so that NO 2 ​ gas is removed and resultant solution is titrated using 2 L , ( 0.8 M ) NaOH solution. Find percentage purity of zinc simple when zinc react with HNO 3 ​ , zinc nitrate, NO 2 ​ and H 2 ​ O are formed. Assume NaOH only reacts with HNO 3 ​ . ( Zn = 65 )

TestHub · Accepted Answer

The initial moles of HNO₃ are calculated from its concentration and volume. The moles of unreacted HNO₃ are determined from the titration with NaOH. The difference gives moles of HNO₃ reacted with Zn. Using the balanced reaction, moles of Zn are found, leading to its mass and percentage purity.

Initial moles of HNO₃ = $(0.2 L) \times (63 g /100 m L) \times (1 m o l /63 g) \times (1000 m L / L) = 2 m o l$

Moles of HNO₃ reacted with NaOH = $(0.8 M) \times (2 L) = 1.6 m o l$

Moles of HNO₃ reacted with Zn = $2 m o l - 1.6 m o l = 0.4 m o l$

The balanced reaction is:

Zn(s) + 4HNO₃(aq) $\to$ Zn(NO₃)₂(aq) + 2NO₂(g) + 2H₂O(l)

From the stoichiometry, 1 mole of Zn reacts with 4 moles of HNO₃.

Moles of Zn = $(0.4 m o l H N O_{3}) \times (1 m o lZ n /4 m o l H N O_{3}) = 0.1 m o lZ n$

Mass of Zn = $(0.1 m o l) \times (65 g / m o l) = 6.5 g$

Percentage purity of Zn = $(6.5 g /10 g) \times 100% = 65%$

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