Question

(Zn + 4HNO_3 Zn(NO_3)_2 + 2H_2O + 2NO_2) 
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) 
 

Hint:

Always check the coefficients in the balanced chemical equation; they tell you the "recipe" or the molar ratio required for the reaction to occur!

Answer 1

(a) Moles of Zinc:
Moles of Zn = Given Mass / Atomic Mass = 32.5 / 65 = 0.5 moles

(b) Mass of Nitric Acid:
From the equation, 1 mole of Zn reacts with 4 moles of HNO3.
Therefore, 0.5 moles of Zn will react with:
0.5 × 4 = 2 moles of HNO3
Molecular weight of HNO3 = 1 + 14 + (16 × 3) = 63 g/mol.
Mass = Moles × Molar Mass = 2 × 63 = 126 g

(c) Volume of NO2 at STP:
From the equation, 1 mole of Zn produces 2 moles of NO2.
Therefore, 0.5 moles of Zn will produce:
0.5 × 2 = 1 mole of NO2
Since 1 mole of any gas occupies 22.4 L at STP:
Volume = 22.4 litres