ICSEClass XChemistryMole Concept and Stoichiometry03 Mark2026 Moderate

QQuestion

Zn + 4HNO3 ⟶ Zn(NO3)2 + 2H2O + 2NO2

32.5 g of zinc reacts with concentrated nitric acid as given in the above equation.

(a) How many moles of zinc was required in the reaction?

(b) Find the mass of nitric acid needed to react with 32.5 g of zinc.

(c) Find the volume of nitrogen dioxide liberated in (b).

[Atomic weight: H=1, N=14, O=16, Zn=65]

Exam-ready answer 03 Mark

Answer

Zn + 4HNO3 ⟶ Zn(NO3)2 + 2H2O + 2NO2

(a) Number of moles of Zn = \frac{\mathrm{Mass}}{\mathrm{Molar\;mass}}=\frac{32.5}{65}=0.5 mole

(b) From the equation, 65 g of Zn reacts with 252 g of HNO3.

∴ 32.5 g of Zn reacts with \frac{252}{65}\times 32.5=126 g of HNO3.

(c) From the equation, 65 g of Zn produces 2 × 22.4 = 44.8 L of NO2 at STP.

∴ 32.5 g of Zn produces \frac{44.8}{65}\times 32.5=22.4 L of NO2 at STP.

Hence, 0.5 mole of zinc is required, 126 g of nitric acid is needed, and 22.4 L of NO2 is liberated at STP.

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